\documentclass[notoc,notitlepage]{tufte-book}
% \nonstopmode % uncomment to enable nonstopmode

\usepackage{classnotetitle}

\title{PMATH467 --- Algebraic Geometry}
\author{Johnson Ng}
\subtitle{Classnotes for Winter 2019}
\credentials{BMath (Hons), Pure Mathematics major, Actuarial Science Minor}
\institution{University of Waterloo}

\usepackage{bm}
\input{latex-classnotes-preamble.tex}

\DeclareMathOperator{\Int}{Int}
\DeclareMathOperator{\Span}{span}

\begin{document}
\input{latex-classnotes-header.tex}

\chapter*{Preface}%
\addcontentsline{toc}{chapter}{Preface}
\label{chp:preface}
% chapter preface

The basic goal of the course is to be able to find \hlnotea{algebraic invariants}, which we shall use to classify topological spaces up to homeomorphism.

Other questions that we shall also look into include a uniqueness problem about manifolds; in particular, how many manifolds exist for a given invariant up to homeomorphism? We shall see that for a \hlnotea{2-manifold}, the only such manifold is the 2-dimensional sphere $S^2$. For a $4$-manifold, it is the $4$-dimensional sphere $S^4$. In fact, for any other $n$-manifold for $n > 4$, the unique manifold is the respective $n$-sphere. The problem is trickier with the $3$-manifold, and it is known as the Poincar\'{e} Conjecture, solved in 2003 by Russian Mathematician \href{https://en.wikipedia.org/wiki/Grigori_Perelman}{Grigori Perelman}. Indeed, the said manifold is homeomorphic to the $3$-sphere.

For this course, you are expected to be familiar with notions from \hlnotea{real analysis},
such as topology, and concepts from \hlnotea{group theory}. Due to the structure of which this
course is designed, each lecture may be much longer than reality, as I am also making heavy
references to the recommended text that is Lee's Introduction to Topological Manifolds
\cite{johnlee2000}.

The following topics shall be covered:
\begin{enumerate}
  \item Point-Set Topology
  \item Introduction to Topological Manifolds
  \item Simplicial complexes \& Introduction to Homology
  \item Fundamental Groups \& Covering Spaces
  \item Classification of Surfaces
\end{enumerate}

\textbf{\newthought{Feb 12th}} I have decided to delve deeper into the recommended text
as the organization of the course demands. As so, changes might be made to earlier
lectures as I go further down, so as to introduce the definitions and provide propositions
at a timing deemed appropriate. To keep track of the changes, please look for
\hlnotea{PMATH467} among the commits on
\url{https://gitlab.com/japorized/TeX_notes/commits/master} using the provided filter. If
you are unfamiliar with version controlling and writing in \LaTeX, once you have found the
lecture that you wish to compare, expand the \hlnotea{diff} for \texttt{classnotes.tex}
if it is collapsed.

\section*{Basic Logistics for the Course}%
% section basic_logistics_for_the_course

I shall leave this here for my own notes, in case something happens to my hard copy.

\begin{itemize}
  \item OH: (Tue) 1630 - 1800, (Fri) 1245 - 1320
  \item OR: MC 6457
  \item EM: aaleyasin
\end{itemize}

% section basic_logistics_for_the_course (end)

% chapter preface (end)

\tuftepart{Point-Set Topology}

\chapter{Lecture 1 Jan 07th}%
\label{chp:lecture_1_jan_07th}
% chapter lecture_1_jan_07th
\nocite{johnlee2000}

We will not be too rigorous in this part.

\section{Euclidean Space}%
\label{sec:euclidean_space}
% section euclidean_space

For any $(x_1, \ldots, x_m) \in \mathbb{R}^m$, we can measure its distance from the origin $0$ using either
\begin{itemize}
  \item $\norm{x}_\infty = \max \{ \abs{ x_i } \}$ (the supremum-norm);
  \item $\norm{x}_2 = \sqrt{ \sum (x_j)^2 }$ (the $2$-norm); or
  \item $\norm{x}_p = \left( \sum \abs{ x_j }^p \right)^\frac{1}{p}$ (the $p$-norm),
\end{itemize}
where we may define a ``distance'' by
\begin{equation*}
  d_p(x, y) = \norm{ x - y }_p.
\end{equation*}

\begin{defn}[Metric]\index{Metric}\label{defn:metric}
  Let $X$ be an arbitrary space. A function $d : X \times X \to \mathbb{R}$ is called a \hlnoteb{metric} if it satisfies
  \begin{enumerate}
    \item (symmetry) $d(x, y) = d(y, x)$ for any $x, y \in X$;
    \item (positive definiteness) $d(x, y) \geq 0$ for any $x, y \in X$, and $d(x, y) = 0 \iff x = y$; and
    \item (triangle inequality) $\forall x, y, z \in X$
      \begin{equation*}
        d(x, y) \leq d(x, z) + d(y, z).
      \end{equation*}
  \end{enumerate}
\end{defn}

\begin{defn}[Open and Closed Sets]\index{Open sets}\index{Closed sets}\label{defn:open_and_closed_sets}
  Given a space $X$ with a metric $d$, and $r > 0$, the set
  \begin{equation*}
    B(x, r) := \{ w \in X \mid d(x, w) < r \}
  \end{equation*}
  is called the \hlnoteb{open ball} of radius $r$ centered at $x$. An \hlnoteb{open set} $A$ is such that $\forall a \in A$, $\exists r > 0$ such that
  \begin{equation*}
    B(a, r) \subseteq A.
  \end{equation*}
  We say that a set is \hlnoteb{closed} if its complement is open.
\end{defn}

\begin{defn}[Continuous Map]\index{Continuous Map}\label{defn:continuous_map}
  A function
  \begin{equation*}
    f : (X, d_1) \to (Y, d_2)
  \end{equation*}
  is said to be \hlnoteb{continuous} if the preimage of an open set in $Y$ is open in $X$.\marginnote{See \href{https://tex.japorized.ink/PMATH351F18/classnotes.pdf\#thm.45}{notes} on Real Analysis for why we defined a continuous map in such a way.}
\end{defn}

\begin{warning}
  This definition does not imply that a continuous map $f$ maps open sets to open sets.
\end{warning}

\begin{defn}[Open and Closed Maps]\index{Open Map}\index{Closed Maps}\label{defn:open_and_closed_maps}
  A mapping $f : X \to Y$ is said to be \hlnoteb{open} if for all open $U \subset X$, $f(U)$ is
  open. We say that $f$ is a \hlnoteb{closed map} if for all closed $F \subset X$, $f(F)$ is
  closed.
\end{defn}
\marginnote{
\begin{ex}\label{ex:homeomorphm_iff_open_and_closed}
  Suppose $f : X \to Y$ is a bijective continuous map. Then TFAE.
  \begin{enumerate}
    \item $f$ is a homeomorphism.
    \item $f$ is open.
    \item $f$ is closed.
  \end{enumerate}
\end{ex}
}

\begin{ex}
  Contruct a function on $[0, 1]$ which assumes all values between its maximum and minimum, but is not continuous.
\end{ex}

\begin{solution}
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw[->] (-0.5, 0) -- (2, 0);
      \draw[->] (0, -0.5) -- (0, 2);
      \draw[dashed] (-0.5, 1) node[left] {$1$} -- (2, 1);
      \draw plot[smooth,domain=0:0.5] ({\x}, {2 * \x});
      \draw plot[smooth,domain=0.5:1] ({\x}, {1 - \x});
    \end{tikzpicture}
    \caption{Function that assumes all values between its extremas, but is not continuous.}\label{fig:function_that_assumes_all_values_between_its_extremas_but_is_not_continuous_}
  \end{marginfigure}
  Consider the piecewise function
  \begin{equation*}
    f(x) = \begin{cases}
      2x & 0 \leq x < \frac{1}{2} \\
      x - \frac{1}{2} & x \geq \frac{1}{2}.
    \end{cases}
  \end{equation*}
  It is clear that the maximum and minimum are $1$ and $0$ respectively, and $f$
  assumes all values between $0$ and $1$. However, the function is
  clearly not continuous, particularly at $\frac{1}{2}$.
\end{solution}

\begin{defn}[Homeomorphism]\index{Homeomorphism}\label{defn:homeomorphism}
  A function $f$ is a \hlnoteb{homeomorphism} if it is a bijection and both $f$ and $f^{-1}$ are continuous.
\end{defn}

\begin{eg}
  The function
  \begin{equation*}
    f : [ 0, 2\pi ) \to \mathbb{R}^2 \text{ given by } \theta \mapsto (\cos \theta, \sin \theta)
  \end{equation*}
  is not homeomorphic, since if we consider an alternating series that converges to $0$ on the unit circle on $\mathbb{R}^2$, we have that the preimage of the series does not converge and $f^{-1}$ is in fact discontinuous.
\end{eg}

% TODO: define convergence; see Introduction to Topological Manifolds by J. Lee pg. 20

\newthought{Now}, we want to talk about topologies without referring to a metric.

\begin{defn}[Topology]\index{Topology}\label{defn:topology}
  Let $X$ be a space. We say that the set $\mathcal{T} \subseteq \mathcal{P}(X)$ is a \hlnoteb{topology} if
  \begin{enumerate}
    \item $X, \emptyset \in \mathcal{T}$;
    \item if $\left\{ x_\alpha \right\}_{\alpha \in A} \subseteq \mathcal{T}$ for an arbitrary index set $A$, then
      \begin{equation*}
        \bigcup_{\alpha \in A} x_\alpha \in \mathcal{T}; \text{ and }
      \end{equation*}
    \item If $\{ x_\beta \}_{\beta \in B} \subset \mathcal{T}$ for some finite index set $B$, then
      \begin{equation*}
        \bigcap_{\beta \in B} x_\beta \in \mathcal{T}.
      \end{equation*}
  \end{enumerate}
\end{defn}

% section euclidean_space (end)

% chapter lecture_1_jan_07th (end)

\chapter{Lecture 2 Jan 09th}%
\label{chp:lecture_2_jan_09th}
% chapter lecture_2_jan_09th

\section{Euclidean Space (Continued)}%
\label{sec:euclidean_space_continued}
% section euclidean_space_continued

In the last lecture, from metric topology, we generalized the notion to a more abstract one
that is based solely on open sets.

\begin{eg}
  Let $X$ be a set. The following two are
  uninteresting examples of topologies:
  \begin{enumerate}
    \item The \hldefn{trivial topology} $\mathcal{T} = \left\{ \emptyset, X \right\}$.
    \item The \hldefn{discrete topology} $\mathcal{T} = \mathcal{P}(X)$.
  \end{enumerate}
\end{eg}

\newthought{We shall now} continue with looking at
more concepts that we shall need down the road.

\begin{defn}[Closure of a Set]\index{Closure}\label{defn:closure_of_a_set}
  Let $A$ be a set. Its \hlnoteb{closure},
  denoted as $\bar{A}$, is defined as
  \begin{equation*}
    \bar{A} = \bigcap_{C \supset A}^{C: \text{ closed }} C.
  \end{equation*}
  It is the smallest closed set that contains $A$.
\end{defn}

\begin{note}
  In metric topology, one typically defines the closure of a set by taking the union of $A$ and its limit points.
\end{note}

\begin{defn}[Interior of a Set]\index{Interior}\label{defn:interior_of_a_set}
  Let $A$ be a set. Its \hlnoteb{interior},
  denoted either as $\text{Int }(A)$, $A^\circ$ or $\Int(A)$,
  is defined as
  \begin{equation*}
    \Int(A) = \bigcup_{G \subseteq A}^{G: \text{ open }} G.
  \end{equation*}
\end{defn}

\begin{defn}[Boundary of a Set]\index{Boundary}\label{defn:boundary_of_a_set}
  Let $A$ be a set. Its \hlnoteb{boundary},
  denoted as $\partial A$, is defined as
  \begin{equation*}
    \partial A = \bar{A} \setminus \Int(A).
  \end{equation*}
\end{defn}

\begin{ex}
  Let $A$ be a set. Prove that $\partial A$ is closed.
\end{ex}

\begin{proof}
  Notice that
  \begin{equation*}
    ( \partial A )^C = ( \bar{A} \setminus \Int(A) )^C = X \setminus \bar{A} \cup \Int(A) = X \cap \bar{A}^C \cup \Int(A)
  \end{equation*}
  which is open.
\end{proof}

\begin{ex}
  Let $A$ be a set. Show that
  \begin{equation*}
    \partial ( \partial A ) = \partial A.
  \end{equation*}
\end{ex}

\begin{proof}
  First, notice that $\Int(\partial A) = \emptyset$.
  Since $\partial A$ is closed, $\bar{\partial A} = \partial A$.
  Then
  \begin{equation*}
    \partial ( \partial A ) = \bar{\partial A} \setminus \Int(\partial A)
      = \partial A \setminus \emptyset = \partial A
  \end{equation*}
\end{proof}

\begin{eg}
  We know that $\mathbb{Q} \subseteq \mathbb{R}$,
  and $\bar{\mathbb{Q}} = \mathbb{R}$.
  We say that $\mathbb{Q}$ is dense in $\mathbb{R}$.
\end{eg}

\begin{defn}[Dense]\index{Dense}\label{defn:dense}
  We say that a subset $A$ of a set $X$ is dense if
  \begin{equation*}
    \bar{A} = X.
  \end{equation*}
\end{defn}

\begin{eg}
  From the last example, we have that $\Int(\mathbb{Q}) = \emptyset$.
\end{eg}

\begin{defn}[Limit Point]\index{Limit Point}\label{defn:limit_point}
  We say that $p \in X \supseteq A$ is a limit point of $A$
  if any neighbourhood of $p$ has a nontrivial intersection
  with $A$.
\end{defn}

\begin{eg}[A Topologist's Circle]\label{eg:a_topologists_circle}
  Consider the function
  \begin{equation*}
    f(x) = \begin{cases}
      \sin \frac{1}{x} & x \neq 0 \\
      0                & x = 0
    \end{cases}
  \end{equation*}
  on the interval $\left[ -\frac{1}{2 \pi}, \frac{1}{2 \pi} \right]$.
  Extend the function on both ends such that we obtain \cref{fig:a_topologists_circle}
  (See also: \href{https://www.desmos.com/calculator/figksl5wmv}{Desmos}).
  \begin{figure}[ht]
    \centering
    \includegraphics[width=0.7\linewidth]{images/topologists-circle.png}
    \caption{A Topologist's Circle}
    \label{fig:a_topologists_circle}
  \end{figure}

  The limit points of the graph includes all the points
  on the straight line from $(0, -1)$ to $(0, 1)$, including the endpoints.
  This is the case because for any of the points on this line, for any
  neighbourhood around the point, the neighbourhood intersects the
  graph $f$ infinitely many times.
\end{eg}

\newthought{Going back to continuity}, given a function $f$,
how do we know if $f^{-1}$ maps an open set to an open set?

We can actually reduce the problem to only looking at open balls.
But why are we allowed to do that?

\begin{defn}[Basis of a Topology]\index{Basis}\label{defn:basis_of_a_topology}
  \marginnote{Note that while the definition is similar to that of a cover, we are now ``covering'' over sets and not points.}
  Given a topology $\mathcal{T}$, we say that
  $\mathcal{B} = \left\{ B_{\alpha} \right\}_{\alpha \in I}$
  is a \hlnoteb{basis} if $\forall T \in \mathcal{T}$, there
  exists $J \subset I$ such that
  \begin{equation*}
    T = \bigcup_{\alpha \in J} B_\alpha.
  \end{equation*}
\end{defn}

\begin{eg}\label{eg:basis_of_r_n}
  Let $\mathcal{T}$ be the Euclidean topology on $\mathbb{R}$.
  Then we can take
  \begin{equation*}
    \mathcal{B} = \left\{ (a, b) \mmid a, b \in \mathbb{R}, \, a \leq b \right\}.
  \end{equation*}
  Note that $\mathcal{B}$ is \hlnotea{uncountable}.
  We can, in fact, have
  \sidenote{Recall from
  \href{https://tex.japorized.ink/PMATH351F18/classnotes.pdf}{PMATH 351}
  that we can write $\mathbb{R}$ as a disjoint union
  of open intervals with rational endpoints.}
  \begin{equation*}
    \mathcal{B}_1 = \left\{ (a, b) \mmid a, b \in \mathbb{Q}, a \leq b \right\},
  \end{equation*}
  which is countable, as a basis for $\mathbb{R}$.
  Furthermore, we can consider the set
  \begin{equation*}
    \mathcal{B}_2 = \left\{ (a, b) \mmid a \leq b, \, a = \frac{m}{2^p}, \, b = \frac{n}{2^q}, \, m, n, p, q \in \mathbb{Z} \right\},
  \end{equation*}
  which is also a countable basis for $\mathbb{R}$.
  Notice that
  \begin{equation*}
    \mathcal{B}_2 \subseteq \mathcal{B}_1 \subseteq \mathcal{B}.
  \end{equation*}
\end{eg}

\begin{eg}
  In $\mathbb{R}^2$, we can do a similar construction of $\mathcal{B}$,
  $\mathcal{B}_1$, and $\mathcal{B}_2$ as in the last example and use
  them as a basis for $\mathbb{R}^2$. In particular, we would have
  \begin{equation*}
    \mathcal{B} = \left\{ (a_1, b_1) \times (a_2, b_2) \mmid a_1, a_2, b_1, b_2 \in \mathbb{R} \right\}.
  \end{equation*}
  This is called a \hldefn{dyadic partitioning} of $\mathbb{R}^2$.
\end{eg}

\begin{eg}
  Let $(X_1, \mathcal{T}_1)$ and $(X_2, \mathcal{T}_2)$ be two topological spaces.
  Then the Cartesian product $X_1 \times X_2$ has topology induced
  from $\mathcal{T}_1$ and $\mathcal{T}_2$ by taking the set
  \begin{equation*}
    \mathcal{B} = \left\{ \beta_1 \times \beta_2 \mmid \beta_1 \in \mathcal{T}_1, \, \beta_2 \in \mathcal{T}_2 \right\}
  \end{equation*}
  as the basis.
\end{eg}

\begin{ex}
  Prove that
  \begin{enumerate}
    \item $\beta_1$ and $\beta_2$ can be taken to be elements of
      bases $\mathcal{B}_1 \subset \mathcal{T}_1$ and
      $\mathcal{B}_2 \subset \mathcal{T}_2$, respectively.
    \item the product topology on $\mathbb{R}^2$ is the same
      as the Euclidean topology.
  \end{enumerate}
\end{ex}

% section euclidean_space_continued (end)

% chapter lecture_2_jan_09th (end)

\chapter{Lecture 3 Jan 11th}%
\label{chp:lecture_3_jan_11th}
% chapter lecture_3_jan_11th

\section{Euclidean Space (Continued 2)}%
\label{sec:euclidean_space_continued_2}
% section euclidean_space_continued_2

\marginnote{The student is recommended to do a quick review for the first 3
chapters of the recommended text.}

Let $\tilde{X}$ be a metric space, and $p, q \in \tilde{X}$ with $p \neq q$.
Then we have that $d(p, q) = r > 0$.
\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \node[circle,fill,inner sep=1pt,label={270:{$p$}}] at (0, 0) {};
    \node[circle,fill,inner sep=1pt,label={270:{$q$}}] at (2, 1) {};
    \draw[latex'-latex'] (0, 0) -- (2, 1) node[midway,fill=background] {r};
    \draw[dashed] (0, 0) circle(0.7);
    \draw[dashed] (2, 1) circle(0.7);
  \end{tikzpicture}
  \caption{Idea of separation}\label{fig:idea_of_separation}
\end{marginfigure}
Then we must have that
\begin{equation*}
  B \left( p, \frac{r}{3} \right) \cap B \left( q, \frac{r}{3} \right) = \emptyset.
\end{equation*}

\begin{ex}
  Prove that the above claim is true. (Use the triangle inequality)
\end{ex}

\begin{proof}
  Suppose $\exists x \in B \left( p, \frac{r}{3} \right) \cap B \left( q, \frac{r}{3} \right)$.
  Then
  \begin{equation*}
    d(p, x) + d(q, x) < \frac{2r}{3} < r = d(p, q),
  \end{equation*}
  which violates the triangle inequality.
\end{proof}

We observe here that the two open sets (or balls) ``separate'' $p$ and $q$.

\begin{defn}[Hausdorff / $T_2$]\index{Hausdorff}\index{$T_2$}\label{defn:hausdorff_t_2_}
  Let $X$ be a topological space. $X$ is said to be \hlnoteb{Hausdorff} or \hlnoteb{$T_2$}
  iff any $2$ distinct points can be separated by disjoint open sets.
\end{defn}

\begin{note}
  \begin{enumerate}
    \item The Hausdorff space (or $T_2$ space) is an important space;
      we can only define a metric on spaces that are $T_2$.
    \item A space is called \hldefn{$T_1$} is for any $p, q \in X$ with $p \neq q$,
      $\exists U \ni p$ open such that $q \notin U$ and $\exists V \ni q$ open
      such that $p \notin v$. It is worth noting that a $T_2$ space is also
      $T_1$.
  \end{enumerate}
\end{note}

\begin{eg}[The Discrete Topology]
  Suppose $X$ is a metric space. For any $x \in X$, we have that $\left\{ x \right\}$ is open.
  Thus for any $x_1, x_2 \in X$, if $x_1 \neq x_2$, then the open sets
  $\left\{ x_1 \right\}$ and $\left\{ x_2 \right\}$ separates $x_1$ and $x_2$.

  This is true as we can define the following metric on the space:
  let $d : X \times X \to \mathbb{R}$ such that
  \begin{equation*}
    d(x_1, x_2) = \begin{cases}
      0 & x_1 = x_2 \\
      1 & x_1 \neq x_2
    \end{cases}
  \end{equation*}
  This topology is called a \hldefn{discrete topology}, and it is a metric space.
\end{eg}

Let $X$ be a metric space and $A \subseteq X$. Then there is a metric induced by
$X$ on $A$, and this in turn induces a topology on $A$.

More generally, if $A \subset X$ where $X$ is some arbitrary topological space,
then a set $U \subseteq A$ is open iff $U = A \cap V$ for some $V \subseteq X$
that is open. In other words, a subset $U$ of $A$ is said to be open iff we can
find an open set $V$ in $X$ such that the intersection of $A$ and $V$ gives us
$U$.

\begin{ex}
  Prove that the construction above gives us a topology.
\end{ex}

\begin{proof}
  Let $A \subseteq X$. We shall show that $\tau_A$ is a topological space induced
  by the topological space $\tau$ of $X$. It is clear that $\emptyset \in \tau_A$,
  since it is open in $X$, and so $A \cap \emptyset = \emptyset$. Since $X$ is open,
  we have $A \cap X = A$, and so $A \in \tau_A$.

  Now if $\left\{ U_\alpha \right\}_{\alpha \in I} \subseteq \tau_A$, then
  $\exists V_\alpha \subseteq X$ such that $U_\alpha = A \cap V_\alpha$. Then
  \begin{equation*}
    \bigcup_{\alpha \in I} U_\alpha = \bigcup_{\alpha \in I} A \cap V_\alpha
      = A \cap \bigcup_{\alpha \in I} V_\alpha,
  \end{equation*}
  and $\bigcup_{\alpha \in I} V_\alpha$ is open in $X$ by the properties of open
  sets. Thus $\bigcup_{\alpha \in I} U_\alpha \in \tau_A$.

  If $\left\{ U_i \right\}_{i = 1}^{n} \subset \tau_A$, then again, by the properties
  of open sets, finite intersection of open sets is open, and so $\bigcap_{i=1}^{n} U_i \in \tau_A$.

\end{proof}

\begin{note}
  We can say the same can be said about closed sets of $A$.
\end{note}

\begin{eg}
  Let $A \subseteq X$ and consider the function
  \begin{equation*}
    \iota_A : A \to X \text{ given by } x \mapsto x,
  \end{equation*}
  which is the inclusion map.

  Then $\iota_A$ is continuous when the topology on $A$ is chosen to be the
  induced subspace topology. This is rather clear; notice that the inverse
  of the inclusion map brings open sets to open sets.
\end{eg}

Let $Y$ be an arbitrary topological space. Then let

\begin{figure}[ht]
  \centering
  \begin{tikzcd}
     &  & X \\
     &  &  \\
    Y \arrow[rr, "f"] \arrow[rruu, "\iota_A \circ f"] &  & A \arrow[uu, "\iota_A"]
  \end{tikzcd}
  \caption{Composition of a function and the inclusion map}
  \label{fig:composition_of_a_function_and_the_inclusion_map}
\end{figure}

where $f$ is continuous. Then $\iota_A \circ f$ is continuous.

\marginnote{
\begin{marginthm}[Characteristic Property of the Subspace Topology]\label{marginthm:characteristic_property_of_the_subspace_topology}
  Suppose $X$ is a topological space and $S \subseteq X$ is a subspace. For any topological space
  $T$, a map $f : Y \to S$ is continuous iff the composite map $\iota_S \circ f : Y \to X$ is
  continuous.
\end{marginthm}
}
\begin{marginfigure}
  \centering
  \begin{tikzcd}
     &  & X \\
     &  &  \\
    Y \arrow[rruu, "\iota_S \circ f"] \arrow[rr, "f"'] &  & S \arrow[uu, "\iota_S"', hook']
  \end{tikzcd}
  \caption{Characteristic Property of the Subspace Topology}
  \label{fig:characteristic_property_of_the_subspace_topology}
\end{marginfigure}

The converse is also true: if $\iota_A \circ f$ is continuous, then $f$ is continuous.
However, we will not prove this. This property is known as the
\hlnoteb{characteristic property of the subspace topology}.

\begin{lemma}[Restriction of a Continuous Map is Continuous]\label{lemma:restriction_of_a_continuous_map_is_continuous}
  Let $X \overset{f}{\to} Y$ be continuous, and $A \subseteq X$
  \sidenote{Here, $A$ is equipped with the subspace topology}. Then
  \begin{equation*}
    f \restriction_{A} : A \to Y
  \end{equation*}
  is also continuous.
\end{lemma}

\begin{propo}[Other Properties of the Subspace Topology]\label{propo:other_properties_of_the_subspace_topology}
  Suppose $S$ is a subspace of the topological space $X$.
  \begin{enumerate}
    \item If $R \subseteq S$ is a subspace of $S$, then $R$ is a subspace of $X$; i.e. the subspace
      topologies that $R$ inherits from $S$ and from $X$ agree.
    \item If $\mathcal{B}$ is a basis for the topology of $X$, then
      \begin{equation*}
        \mathcal{B}_S = \{ B \cap S : B \in \mathcal{B} \}
      \end{equation*}
      is a basis for the topology of $S$.
    \item If $\{p_i\}$ is a sequence of points in $S$ and $p \in S$, then $p_i$ converges to $p$
      in $S$ iff $p_i$ converges to $p$ in $X$.
    \item Every subspace of a Hausdorff space is Hausdorff.
  \end{enumerate}
\end{propo}

% section euclidean_space_continued_2 (end)

\section{Connected Spaces}%
\label{sec:connected_spaces}
% section connected_spaces

Consider the real line $\mathbb{R}$, and consider two disjoint intervals on $\mathbb{R}$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
    \draw[->] (-2, 0) -- (2, 0);
    \draw[-{Arc Barb[length=5pt,width=25pt]},blue] (-2,0) -- (-0.2,0) node[midway,below=15pt] {$U$};
    \draw[{Arc Barb[length=5pt,width=25pt]}-,blue] (0.2,0) -- (2,0) node[midway,below=15pt] {$V$};
    \draw[{Arc Barb[length=5pt,width=15pt]}-{Bracket[length=5pt,width=15pt]},red] (-1.5,0) -- (-0.5,0) node[midway,below] {$A_1$};
    \draw[{Arc Barb[length=5pt,width=15pt]}-{Bracket[length=5pt,width=15pt]},red] (0.5,0) -- (1.5,0) node[midway,below] {$A_2$};
  \end{tikzpicture}
  \caption{Motivation for Connectedness}
  \label{fig:motivation_for_connectedness}
\end{figure}

Observe that we may find two open subsets $U$ and $V$ of $\mathbb{R}$ such that $A_1 \subseteq U$
and $A_2 \subseteq V$, which effectively separates the two intervals on the space $\mathbb{R}$.

\begin{defn}[Disconnectedness]\index{Disconnectedness}\label{defn:disconnectedness}
  A space $X$ is said to be \hlnoteb{disconnected} iff $X$ can be written as a
  disjoint union
  \begin{equation*}
    X = A_1 \coprod A_2
  \end{equation*}
  where $A_1, A_2 \subseteq X$, $A_1 = A_2^C$, that they are both non-empty and open
  \sidenote{It goes without saying that the two sets are also simultaneously closed.}.
\end{defn}

\begin{defn}[Connectedness]\index{Connectedness}\label{defn:connectedness}
  A space $X$ is said to be \hlnoteb{connected} if it is not disconnected.
\end{defn}

\begin{note}
  By the above definitions, we have that $X$ is connected iff for any partition
  $X = A \coprod A^C$ with $A$ being open, either $A$ is $\emptyset$ or $A$ is $X$.
\end{note}

\begin{eg}
  The space $\mathbb{R} \setminus \left\{ 0 \right\}$ is disconnected; our disjoint sets
  are $(-\infty, 0)$ and $(0, \infty)$.
\end{eg}

However, $\mathbb{R}^2 \setminus \left\{ 0 \right\}$ is connected, but it is not easy to
describe why.

\begin{defn}[Path]\index{Path}\label{defn:path}
  A \hlnoteb{path} in a space $X$ from $p$ to $q$ (both in $X$) is a continuous map
  $f : [0, 1] \to X$ such that $f(0) = p$ and $f(1) = q$. We say that $X$ is
  \hldefn{path connected} if $\forall p, q \in X$, there is a path in $X$ from $p$
  to $q$.
\end{defn}

\begin{lemma}[Path Connectedness implies Connectedness]\label{lemma:path_connectedness_implies_connectedness}
  If a space $X$ is path connected, then it is connected.
\end{lemma}

\begin{thm}[From Connected Space to Connected Space]\label{thm:from_connected_space_to_connected_space}
  If $X \overset{f}{\to} Y$ is continuous and $X$ is connected, then
  $\Img(f)$ is connected.
\end{thm}

% section connected_spaces (end)

% chapter lecture_3_jan_11th (end)

\chapter{Lecture 4 Jan 14th}%
\label{chp:lecture_4_jan_14th}
% chapter lecture_4_jan_14th

\section{Connected Spaces (Continued)}%
\label{sec:connected_spaces_continued}
% section connected_spaces_continued

\begin{defn}[Locally Connected]\index{Locally Connected}\label{defn:locally_connected}
  We say that $X$ is \hlnoteb{locally connected} at $x$ if for every open set $V$
  containing $x$ there exists a connected, open set $U$ with $x \in U \subseteq V$. We say that
  the space $X$ is \hlnoteb{locally connected} if it is locally connected $\forall x \in X$.
\end{defn}

\begin{eg}
  The space $S$ generated by the function $\sin \frac{1}{x}$ with $0$ at $x = 0$,
  on the $\mathbb{R}^2$ is not locally connected: consider $(0, y) \in S, y \neq 0$.
  Then any small open ball at this point will contain infinitely many line segments from $S$.
  This cannot be connected, as each one of these constitutes a component,
  within the neighborhood.
\end{eg}

\begin{defn}[Connected Component]\index{Connected Component}\label{defn:connected_component}
  The maximal connected subsets of any topological space $X$ are called
  \hlnoteb{connected components} of the space. The components form a partition
  of the space.
\end{defn}

% section connected_spaces_continued (end)

\section{Compactness}%
\label{sec:compactness}
% section compactness

\begin{defn}[Sequential Compactness]\index{Sequential Compactness}\label{defn:sequential_compactness}
  For $A \subseteq X$, where $X$ is a topological space, if
  $\left\{ x_i \right\}_{i \in I} \subseteq A$, arbitrary sequence in $A$, has a
  convergent subsequence, we say that $A$ is \hlnoteb{sequentially compact}.
\end{defn}

\begin{defn}[Compactness]\index{Compactness}\label{defn:compactness}
  We say that a topological space $X$ is \hlnoteb{compact} if every \hlnotea{open cover}
  of $X$ has a finite \hlnotea{subcover}.
\end{defn}

\begin{lemma}[Compactness implies Sequential Compactness]\label{lemma:compactness_implies_sequential_compactness}
  Compactness implies sequential compactness.
\end{lemma}

\begin{eg}
  $[ 0, 1 )$ is not compact: consider the open cover
  $\left\{ \left[ 0, 1 - \frac{1}{n} \right) \right\}_{n \in \mathbb{N}}$, which contains
  $[0, 1)$ as $n \to \infty$. But whenever $n$ is finite, we have $1 - \frac{1}{n} < 1$,
  and so any finite collection of the $\left[0, \frac{1}{n}\right)$ is not a cover of
  $[0, 1)$.
\end{eg}

\begin{thm}[Continuous Maps map Compact Sets to Compact Images]\label{thm:continuous_maps_map_compact_sets_to_compact_images}
  Let $f : X \to Y$ be continuous, where $X$ is compact. Then $f(X)$ is compact.
\end{thm}

\begin{proof}
  Let $\left\{ U_\alpha \right\}_{\alpha \in I}$ be an open cover of $f(X)$. Since
  $f$ is continuous, we have that $f^{-1} \left( U_\alpha \right)$ is open for each
  $\alpha \in I$. Since $f$ is bijective between the image set and its domain, we
  have that $\left\{ f^{-1} \left( U_\alpha \right) \right\}_{\alpha \in I}$ is an
  open cover of $X$. Since $X$ is compact, this cover has a finite subcover, say
  $\left\{ f^{-1} \left( U_i \right) \right\}_{i = 1}^{n}$. Thus
  \begin{equation*}
    X = \bigcup_{i=1}^{n} f^{-1} \left( U_i \right).
  \end{equation*}
  Thus
  \begin{equation*}
    f(X) = f \left( \bigcup_{i=1}^{n} f^{-1} \left( U_i \right) \right) = \bigcup_{i=1}^{n} U_i.
  \end{equation*}
  Hence $\left\{ U_\alpha \right\}_{\alpha \in I}$ has a finite subcover and so
  $f(X)$ is compact.
\end{proof}

\begin{crly}[Homeomorphic Maps map Compact Sets to Compact Sets]\label{crly:homeomorphic_maps_map_compact_sets_to_compact_sets}
  Let $X \overset{f}{\to} Y$ be a homeomorphism. Then $X$ is compact iff
  $Y$ is compact.
\end{crly}

\begin{note}
  Compactness is a topological property.
\end{note}

\begin{lemma}[Properties of Compact Sets]\label{lemma:properties_of_compact_sets}
  \begin{enumerate}
    \item A closed subset of a compact space is compact.
    \item A compact subset of a topological space is closed provided that the
      space is Hausdorff.
    \item In a metric space, a compact set is bounded.
    \item Finite (Cartesian) product of compact sets is compact.
  \end{enumerate}
\end{lemma}

The proof for the first item is simple: consider an open cover of the closed subset,
and union them with the complement of the closed subset. This covers the entire space,
and so it has a finite subcover. We just need to then remove that complement set, and
that would be a finite subcover for the closed subset.

\begin{eg}
  The subset $[-a, b]$, $a, b \in \mathbb{R}$, is compact.
\end{eg}

\begin{eg}
  $[0, 1]^{\mathbb{N}}$ is not compact: the space is equivalent to $\ell_\infty$.
\end{eg}

\begin{thm}[Heine-Borel]\index{Heine-Borel}\label{thm:heine_borel}
  Let $X \subseteq \mathbb{R}^n$. Then $X$ is compact iff $X$ is closed and bounded.
\end{thm}

\begin{proof}
  \hlbnoted{$(\implies)$} We say that compactness impliess boundedness. Also, since
  $\mathbb{R}^n$ is Hausdorff, $X$ is closed.\sidenote{Both from
  \cref{lemma:properties_of_compact_sets}}.

  \noindent
  \hlbnoted{$(\impliedby)$} $X$ is bounded implies that $X \subseteq [-R, R]^n$ with $R$
  sufficiently large. Since $X$ is closed, and $[-R, R]^n$ is compact, $X$ is necessarily
  compact by \cref{lemma:properties_of_compact_sets}.
\end{proof}

\begin{thm}[Bolzano-Weierstrass]\index{Bolzano-Weierstrass}\label{thm:bolzano_weierstrass}
  Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.
\end{thm}

\begin{ex}
  Prove \cref{thm:bolzano_weierstrass} as an exercise.
\end{ex}

% section compactness (end)

We shall start the next part this lecture.

\section{Manifolds}%
\label{sec:manifolds}
% section manifolds

\begin{defn}[Locally Homeomorphic]\index{Locally Homeomorphic}\label{defn:locally_homeomorphic}
  A space is said to be \hlnoteb{locally homeomorphic} to $\mathbb{R}^n$ provided that
  $\forall x \in X$, $\exists U \ni x$ open such that $U$ is homeomorphic to $\mathbb{R}^n$.
\end{defn}

\begin{defn}[Manifold]\index{Manifold}\label{defn:manifold}
  An $n$-dimensional \hlnoteb{manifold} is a \hlbnotec{second countable}\sidenote{
  a topological space is said to be \hldefn{second countable} if its basis is
  countable. Note that for a second countable set $X$, every open cover of $X$ has a countable
  subcover (see pg 32 of \citet{johnlee2000}).}, \hlbnotec{Hausdorff} topological space that is
  \hlbnotec{locally homeomorphic} to $\mathbb{R}^n$.
\end{defn}
\marginnote{
\begin{procedure}[Show that $X$ is a manifold]
  To show that a space is an $n$-dimensional manifold, check that the space is
  \begin{enumerate}
    \item Hausdorff;
    \item locally Euclidean; and
    \item locally homeomorphic to $\mathbb{R}^n$.
  \end{enumerate}
  This section covers some tools that helps make the process easier.
\end{procedure}}

\begin{note}
  We may also call the last criterion of being a manifold, that is, to be locally
  homeomorphic to $\mathbb{R}^n$, as to be \hldefn{locally Euclidean} of dimension
  $n$.
\end{note}

\begin{note}
  One can give an equivalent definition of locally homeomorphic by requiring that
  $U$ be homeomorphic to an open ball $B^n \subseteq \mathbb{R}^n$. Notice that
  $B^n$ is homeomorphic to $\mathbb{R}^n$ \sidenote{By scaling, really.}
\end{note}

The following is a quick fact about second countable spaces, which will be helpful
when we start creating subspaces from manifolds.

\begin{propo}[Subspaces of Second Countable Spaces are Second Countable]\label{propo:subspaces_of_second_countable_spaces_are_second_countable}
  Every subspace of a second countable space is second countable.
\end{propo}

\begin{eg}
  Let $B^n = B^n(0, 1) \subseteq \mathbb{R}^n$. Then $B^n$ is homeomorphic to
  $\mathbb{R}^n$.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw[dashed] (0, 0) circle(2);
      \draw[dashed] (0.5, 0.5) circle(0.5);
      \node[circle,fill,inner sep=1pt] at (0.5, 0.5) {};
    \end{tikzpicture}
    \caption{Open ball in an open set in $\mathbb{R}^2$}\label{fig:open_ball_in_an_open_set_in_r_2_}
  \end{marginfigure}
\end{eg}

\begin{eg}
  Now consider the closed ball $\bar{B}^n = \bar{B}^n(0, 1) \subset \mathbb{R}^n$. This is
  actually not a manifold, but we are not yet there to prove this. This sort of a structure
  motivates us to the next definition.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (0, 0) circle(1);
      \node[circle,fill,inner sep=1pt] at (-0.7071067811865475244, 0.7071067811865475244) {};
      \clip (0, 0) circle(1);
      \draw[dashed] (-0.7071067811865475244,0.7071067811865475244) circle(0.5);
    \end{tikzpicture}
    \caption{Open ball on a point on the boundary of a closed set}\label{fig:open_ball_on_a_point_on_the_boundary_of_a_closed_set}
  \end{marginfigure}
\end{eg}

\begin{defn}[Manifold with Boundary]\index{Manifold with Boundary}\label{defn:manifold_with_boundary}
  An $n$-dimensional space that is second countable and Hausdorff, such that $\forall x \in X$,
  there exists a neighbourhood either homeomorphic to $\bar{B}^n \subseteq \mathbb{R}^n$ or
  $B^n \cap \mathbb{H}^n$.
\end{defn}

\begin{note}
  Note that $\mathbb{H}^n$ is defined as
  \begin{equation*}
    \mathbb{H}^n = \left\{ ( x_1, \ldots, x_n ) : x_n \geq 0 \right\}.
  \end{equation*}
  For instance, $\mathbb{H}^2$ has the following graph:
  \begin{figure}[ht]
    \centering
    \begin{tikzpicture}
      \draw[->] (-2, 0) -- (2, 0);
      \draw[->] (0, -0.5) -- (0, 2);
	    \draw[line width=0pt,draw=none,fill=blue,fill opacity=0.25] (-2,0)--(2,0)--(2,2)--(-2,2);
    \end{tikzpicture}
    \caption{Graph of $\mathbb{H}^2$}
    \label{fig:graph_of_h_2}
  \end{figure}
\end{note}

% section manifolds (end)

% chapter lecture_4_jan_14th (end)

\tuftepart{Introduction to Topological Manifolds}%
\label{prt:introduction_to_topological_manifolds}

\chapter{Lecture 5 Jan 16th}%
\label{chp:lecture_5_jan_16th}
% chapter lecture_5_jan_16th

\section{Manifolds (Continued)}%
\label{sec:manifolds_continued}
% section manifolds_continued

\begin{defn}[Interior Point]\index{Interior Point}\label{defn:interior_point}
  A point $x \in M$ is called an \hlnoteb{interior point} if there is a local
  homeomorphism
  \begin{equation*}
    \phi : \mathcal{U} \to \mathbb{B} \subseteq \mathbb{R}^n,
  \end{equation*}
  where $\mathcal{U}$ is open.
\end{defn}

In the last lecture we asked ourselves the following: how do we know if the
idea of `being on a boundary' is a well-defined notion? In particular, how
do well tell the difference between the following two graphs, mathematically?

\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \draw[->] (-2, 0) -- (2,0);
    \draw[->] (0, -2) -- (0,2);
	  \draw[line width=0pt,draw=none,fill=blue,fill opacity=0.25] (-2,-2)--(2,-2)--(2,2)--(-2,2);
    \node at (1.5,1.5) {$\mathbb{R}^n$};
    \draw[draw=red] (0, 0) circle(1);
    \node[circle,fill=red,inner sep=2pt] at (0,0) {};

    \draw[->] (4, -1) -- (8, -1);
    \draw[->] (6, -2) -- (6, 2);
	  \draw[line width=0pt,draw=none,fill=blue,fill opacity=0.25] (4,-1)--(8,-1)--(8,2)--(4,2);
    \node at (7.5, 1.5) {$\mathbb{H}^n$};
    \draw[draw=red] (5, -1) -- (7, -1) arc(0:180:1) --cycle;
    \node[circle,fill=red,inner sep=2pt] at (6,-1) {};
  \end{tikzpicture}
  \caption{How do we tell the difference between the two graphs?}
  \label{fig:r_n_vs_h_n}
\end{figure*}

\begin{defn}[Boundary Point]\index{Boundary Point}\label{defn:boundary_point}
  A point $x$ is on the boundary of $M$, denoted as $x \in \partial M$, if
  there exists $\mathcal{U} \ni x$ that is open, and a homeomorphism
  \begin{equation*}
    \psi : \mathcal{U} \to \mathbb{B}_{0, 1} \cap \mathbb{H}^n.
  \end{equation*}
\end{defn}

\begin{note}
  Note that the definition of a boundary and interior is different from our earlier
  definitions for the same terminologies. A manifold with boundary will always have
  an empty boundary, as a subset, despite the fact that its boundary as a manifold
  may not be empty.
\end{note}

\begin{note}
  The $\phi$ in \cref{defn:interior_point} and $\psi$ in \cref{defn:boundary_point}
  are called \hldefn{local charts}.

  Also, our definitions do not rule out, e.g.
  \begin{gather*}
    \phi_2 : \mathcal{U} \to \mathbb{B}^2 \subseteq \mathbb{R}^2 \\
    \phi_5 : \mathcal{U} \to \mathbb{B}^5 \subseteq \mathbb{R}^5.
  \end{gather*}
\end{note}

\begin{defn}[Coordinate Chart]\index{Coordinate Chart}\label{defn:coordinate_chart}
  If $M$ is locally homeomorphic to $\mathbb{R}^n$, a homeomorphism from an open subset
  $U \subset M$ to an open subset of $\mathbb{R}^n$ is called a \hlnoteb{coordinate chart}
  (or simply a \hldefn{chart}).
\end{defn}

We shall later on prove that a point cannot simultaneously be a boundary point and
an interior point. This property is called \hlnotea{the invariance of the boundary}.
\sidenote{This also means that $\Int(M) \cap \partial M = \emptyset$. Also, by
\cref{defn:manifold}, $\Int(M) \cup \partial M = M$.}

\begin{note}
  $\Int(M)$ is open. Thus, we can use the same definition about open sets as
  before using $\phi$.
\end{note}

\begin{note}
  In contrast, $\partial M$ is closed; thanks to the invariance of the boundary
  we have that $\partial M = M \setminus \Int(M)$ and $\Int(M)$ is open, and so
  $\left( \partial M \right)^C = \Int(M)$.
\end{note}

We shall also prove the following theorem later on:

\begin{thmnonum}[Invariance of the Dimension]\label{thmnonum:invariance_of_the_dimension}
  The $n$ in $\mathbb{R}^n$ is well-defined.
\end{thmnonum}

\begin{eg}
  Consider the equation
  \begin{equation}\label{eg:equation_of_a_3d_cone}
    x^2 - y^2 - z^2 = 0.
  \end{equation}
  Note that we may write
  \begin{equation*}
    x = \pm \sqrt{ y^2 + z^2 },
  \end{equation*}
  and so the graph generated by \cref{eg:equation_of_a_3d_cone} is as shown in
  \cref{fig:a_3d_cone_in_r_3_}.
  \begin{marginfigure}
    \centering
    \includegraphics[width=\marginparwidth]{images/3d_cone.png}
    \caption{A 3D cone in $\mathbb{R}^3$, from WolframAlpha}\label{fig:a_3d_cone_in_r_3_}
  \end{marginfigure}
  However, this is not a manifold: if we assume that a ball arounnd the origin
  is homeomorphic to $\mathbb{R}^2$, then by removing the point at the origin
  in the cone, the cone becomes two disconnected components, but the ball in
  $\mathbb{R}^2$ homeomorphic to the aforementioned ball is still connected.
\end{eg}

\begin{note}
  An open subset of a manifold is a manifold, by restriction.
\end{note}

\subsection{The $1$-Sphere $S^1$}%
\label{sub:the_1_sphere_s_1_}
% subsection the_1_sphere_s_1_

From \cref{eg:basis_of_r_n}, we have
\begin{itemize}
  \item $[0, 1) \simeq [0, \infty)$ is a manifold with boundary;
  \item $(0, 1) \simeq \mathbb{R}$ is a manifold; and
  \item $[0, 1]$ is a manifold with boundary.
\end{itemize}

\begin{eg}[$S^1$ is a manifold]
  Consider the function $f : [0, 2 \pi) \to e^{i \theta}$. The image of $f$ is
  \begin{figure}[ht]
    \centering
    \begin{tikzpicture}
      \draw[{Bracket[length=5pt,width=20pt]}-{Arc Barb[length=5pt,width=20pt]}] (0, 0) arc(0:360:1.5) node[circle,fill,inner sep=1pt] {};
    \end{tikzpicture}
    \caption{$S^1$ as a manifold}
    \label{fig:s_1_as_a_manifold}
  \end{figure}
  Consider the following two functions $(0, 2\pi) \to \mathbb{C}^2$ by
  \begin{gather*}
    \theta_1 \to e^{i \theta_1} \text{ and } \theta_2 \to e^{i \theta_2 + \pi},
  \end{gather*}
  which gives us the graphs:
  \begin{figure}[ht]
    \centering
    \begin{tikzpicture}
      \draw (0, 0) circle(1);
      \node[circle,draw,inner sep=1.5pt] at (1, 0) {};
      \draw (3, 0) circle(1);
      \node[circle,draw,inner sep=1.5pt] at (2, 0) {};
    \end{tikzpicture}
    \caption{Basis for $S^1$}
    \label{fig:basis_for__s_1_}
  \end{figure}

  \noindent
  respectively. Note that the image of both these functions are not compact.
  Regardless, this gives us a basis for $S^1$, which we notice is countable,
  Hausdorff, and locally homeomorphic to $\mathbb{R}^2$.
\end{eg}

\begin{thm}[$1$-Dimensional Manifolds Determined by Its Compactness]\label{thm:_1_dimensional_manifolds_determined_by_its_compactness}
  Let $M$ be a connected component of a $1$-dimensional manifold. Then either
  \begin{enumerate}
    \item $M$ is compact, in which case if it is
      \begin{itemize}
        \item without a boundary, then $M$ is homeomorphic to $S^1$.
        \item with a boundary, then $M$ is homeomorphic to $[0, 1]$.
      \end{itemize}
    \item $M$ is not compact, in which case if it is
      \begin{itemize}
        \item without a boundary, then $M$ is homeomorphic to $[0, 1)$.
        \item with a boundary, then $M$ is homeomorphic to $(0, 1)$.
      \end{itemize}
  \end{enumerate}
\end{thm}

% subsection the_1_sphere_s_1_ (end)

% section manifolds_continued (end)

% chapter lecture_5_jan_16th (end)

\chapter{Lecture 6 Jan 18th}%
\label{chp:lecture_6_jan_18th}
% chapter lecture_6_jan_18th

\section{Manifolds (Continued 2)}%
\label{sec:manifolds_continued_2}
% section manifolds_continued_2

\subsection{The $1$-Sphere $S^1$ (Continued)}%
\label{sub:the_1_sphere_s_1_continued}
% subsection the_1_sphere_s_1_continued

\paragraph{Set theoretic view of $S^1$}
We showed that $S^1$ is a manifold. We can, in fact, set theoretically, look
at $S^1$ as $A = [0, 1]$ glued at the endpoints, i.e. we \hlnotea{identify}
the points $0$ and $1$ as `the same', and label this notion as $0 \sim 1$.

\paragraph{Topological view of $S^1$}
Topologically, for $0 \sim 1$ in $A$, we can construct an open set around the
point such that the open set is properly contained in $A$.

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
    \draw[{Bracket[length=5pt,width=15pt]}-{Bracket[length=5pt,width=15pt]}] (-2, 0) -- (2, 0);
    \draw[-{Arc Barb[length=5pt,width=15pt]},blue] (-2,0) -- (-1,0);
    \draw[{Arc Barb[length=5pt,width=15pt]}-,blue] (1,0) -- (2,0);
    \node[circle,fill=blue,inner sep=1.5pt] at (-2, 0) {};
    \node[circle,fill=blue,inner sep=1.5pt] at (2, 0) {};
  \end{tikzpicture}
  \caption{Topological representation of $A$}
  \label{fig:topological_representation_of_a}
\end{figure}

But how can we describe this notion mathematically so?

\newthought{Consider} the real line as follows:

\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}[scale=1.5]
    \draw[->] (-5, 0) -- (5, 0);
    \foreach \x in {-4,-3,...,4} {
      \draw[dashed,color=blue] (\x,0.5) -- (\x,-0.5);
      \node[circle,fill,inner sep=2pt,label={270:{$\x$}}] at (\x, 0) {};
      \node[label={[label distance=1]above:{$T^{\x}(a)$}}] at (\x + 0.3, 0) {$( \times )$};
      \node at (\x + 0.7, 0) {$\blacktriangle$};
    }
  \end{tikzpicture}
  \caption{Breaking down the real line into parts}
  \label{fig:breaking_down_the_real_line_into_parts}
\end{figure*}

Let's define $T : \mathbb{R} \to \mathbb{R}$ such that $x \mapsto x + 1$. Clearly
so, $T$ is bijective and, in particular, has an inverse $x \mapsto x - 1$.

Notice that within each interval $[x, x + 1]$, we can find a $\times$ and
$\blacktriangle$ at the same distance from $x$. Also, notice that we can use the
same radius for $\times$ such that the open ball around $\times$ sits in $[x, x + 1]$
for each $x \in \mathbb{Z}$.

Thus, instead of studying the entire real line at once, we can reduce our attention
only to $[0, 1]$, and simply scale the interval with a `scalar multiplication' to
get to wherever we want on the real line.

Now let
\begin{equation*}
  G := \left\{ T^k \mmid k \in \mathbb{Z} \right\},
\end{equation*}
which is evidently a \hldefn{group}. Furthermore, every element in $G$ is a
homeomorphism to $\mathbb{R}$. Let $G$ \hldefn{act on} $\mathbb{R}$, and for
$a \in \mathbb{R}$, consider the \hldefn{orbit} of $a$, which is denoted as
\begin{equation*}
  G \cdot a := \left\{ T^k (a) \mmid k \in \mathbb{Z} \right\}.
\end{equation*}
Then
\begin{equation*}
  S^1 \simeq \text{ the space of all orbits of } G \text{ acting on } \mathbb{R} =: \mathbb{R} / G,
\end{equation*}
where $\simeq$ represents \hyperref[defn:homeomorphism]{homeomorphism}.
\sidenote{$\mathbb{R} / G$ is the \hldefn{quotient space} of $\mathbb{R}$ over $G$.}
Also, notice that here, $G$ is effectively $\mathbb{Z}$.

This realization implies the existence of some topology on $S^1$.

\newthought{Thus far}, we have seen that we may look at $S^1$
\begin{itemize}
  \item set theoretically: as $A = [0, 1]$ with glued endpoints; and
  \item topologically: as $\mathbb{R} \setminus \mathbb{Z}$.
\end{itemize}

\paragraph{$S^1$ as a topological group}
Since $\mathbb{C} \simeq \mathbb{R}^2$, we may think of $S^1$ as a
sphere on the complex plane.
\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}[scale=1.3]
    \draw[->] (-2,0) -- (2, 0);
    \draw[->] (0,-2) -- (0, 2);
    \draw (0, 0) circle(1);
    \draw (0, 0) -- (0.7071067811865475244,0.7071067811865475244)
      node[circle,fill,inner sep=1pt,label={45:{$z_1 = e^{i \theta_1}$}}] {};
    \draw (0, 0) -- (-0.7071067811865475244,0.7071067811865475244)
      node[circle,fill,inner sep=1pt,label={135:{$z_2 = e^{i \theta_2}$}}] {};
    \draw[dashed] (0, 0) -- (-0.7071067811865475244,-0.7071067811865475244)
      node[circle,fill,inner sep=1pt,label={225:{$z_1 z_2 = e^{i ( \theta_1 + \theta_2 )}$}}] {};
    \node[scale=1.3,right] at (1, 1.5) {$S^1 = \left\{ z \in \mathbb{C} \mid \abs{ z } = 1 \right\} = \left\{ e^{i \theta} \mid \theta \in \mathbb{R} \right\}$};
  \end{tikzpicture}
  \caption{$S^1$ on the complex plane}
  \label{fig:_s_1_on_the_complex_plane}
\end{figure*}
We see that this `group' takes on the operation of adding the indices
of the exponents. Thus $G = ( S^1, \cdot )$. Notice that $G$ is indeed
a group equipped with said operation, and for each $z_1 \in G$, there
exists $z_1^{-1} = \frac{1}{z_1} \in G$ such that
$z_1 \cdot \frac{1}{z_1} = 1$.

Furthermore, the function
\begin{gather*}
  \iota : S^1 \to S^1 \text{ given by } z \mapsto \frac{1}{z} \\
  \text{ which is } e^{i \theta} \mapsto e^{- i \theta}
\end{gather*}
is continuous.

Also, the function
\begin{equation*}
  P : S^1 \times S^1 \to S^1 \text{ given by } (z_1, z_2) \mapsto z_1 z_2
\end{equation*}
is continuous.

\begin{defn}[Topological Group]\index{Topological Group}\label{defn:topological_group}
  If $G$ is a group, and functions $\iota$ and $P$ as defined above,
  if both $\iota$ and $P$ are continuous, then we say that $G$ is a
  \hlnoteb{topological group}.
\end{defn}

\paragraph{$S^1$ as a moduli space}

\begin{defn}[Moduli Space]\index{Moduli Space}\label{defn:moduli_space}
  A \hlnoteb{moduli space} is the space of all lines passing through
  the origin.
\end{defn}

On $\mathbb{R}^2$, we have
\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
    \draw[->] (-2, 0) -- (2, 0);
    \draw[->] (0, -2) -- (0, 2);
    \draw (0, 0) circle(1);

    \clip (-2, -2) rectangle (2, 2);
    \foreach \m in {-3,-2.5,...,3} {
      \draw[color=blue] plot ({\x}, {\m * \x});
    }
  \end{tikzpicture}
  \caption{The moduli space on $\mathbb{R}^2$}
  \label{fig:the_moduli_space_on_r_2_}
\end{figure}

First, how can we understand `closeness' in a moduli space?
We can actually look at the difference in the radians of each
line, or really just $x / 360$ and compare the $x$'s.

Also, notice that each line passes through $S^1$ twice. We can
indeed avoid this problem by shifting $S^1$ to one side, as
shown in \cref{fig:shifted_s_1_for_the_moduli_space}.
\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \draw[->] (-1, 0) -- (3, 0);
    \draw[->] (0, -2) -- (0, 2);
    \draw (1, 0) circle(1);

    \clip (-2, -2) rectangle (3, 2);
    \foreach \m in {-3,-2.5,...,3} {
      \draw[color=blue] plot ({\x}, {\m * \x});
    }
  \end{tikzpicture}
  \caption{Shifted $S^1$ for the moduli space}\label{fig:shifted_s_1_for_the_moduli_space}
\end{marginfigure}
Now each line intersects $S^1$ at the origin and another
point on $S^1$, and this intersection is in fact unique.

\subsubsection{The space of $S^1 \times S^1$}%
\label{ssub:the_space_of_s_1_times_s_1_}
% subsubsection the_space_of_s_1_times_s_1_

Observe that the product of two manifolds is a manifold
\sidenote{This is called a \hldefn{product space}.}.

\newthought{Consider} using the set theoretical viewpoint,
with credits to \hlnotea{Felix Klein}, the following figure
\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \draw[{Bracket[length=5pt,width=15pt]}-{Bracket[length=5pt,width=15pt]}] (0, 0) -- (2, 0);
    \draw[{Bracket[length=5pt,width=15pt]}-{Bracket[length=5pt,width=15pt]}] (0, 0) -- (0, 2);
    \node[circle,fill,inner sep=2pt] at (0, 0) {};
    \node[circle,fill,inner sep=2pt] at (2, 0) {};
    \draw (0, 2) -- (2, 2) -- (2, 0);
    \node at (1, 0) {$>$};
    \node at (1, 2) {$>$};
    \node[rotate=90] at (0, 1) {$>>$};
    \node[rotate=90] at (2, 1) {$>>$};

    \node at (3, 1) {$\leadsto$};

    \draw (4, 0.5) arc(-90:90:0.25 and 0.5)
      node[midway,rotate=90] {$>>$};
    \draw (4, 0.5) arc(-90:-270:0.25 and 0.5);
    \draw (4, 1.5) -- (6, 1.5) node[midway] {$>$};
    \draw (4, 0.5) -- (6, 0.5) node[midway] {$>$};
    \draw (6, 0.5) arc(-90:90:0.25 and 0.5)
      node[midway,rotate=90] {$>>$};
    \draw[dashed] (6, 0.5) arc(-90:-270:0.25 and 0.5);

    \node at (7, 1) {$\leadsto$};

    \draw (8.3,1) to[bend left] (8.8,1);
    \draw (8.1,1.1) to[bend right] (9,1.1);
    \draw[rotate=0] (8.5,1) ellipse (30pt and 15pt);
  \end{tikzpicture}
  \caption{$S^1 \times S^1$ becomes a cylinder by identifying the edges}
  \label{fig:_s_1_times_s_1_becomes_a_cylinder_by_identifying_the_edges}
\end{figure*}
By joining the sides with $>$, where we identify the endpoints,
we can go from the figure introduced by Klein to a cylinder. Then
by identifying the sides with $>>$, we get a \hldefn{torus}.

\newthought{Now on} a Cartesian plane, observe that
\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \foreach \z in {-2,-1,...,3} {
      \draw[dashed,color=blue!50!foreground] (\z, -2) -- (\z, 2);
    }
    \draw[->] (-2, 0) -- (3, 0);
    \draw[->] (0, -2) -- (0, 2);
    \draw (0, 0) -- (1, 0) -- (1, 1) -- (0, 1) --cycle;
    \node at (0.5, 0) {$>$};
    \node at (0.5, 1) {$>$};
    \node[rotate=90] at (0, 0.5) {$>>$};
    \node[rotate=90] at (1, 0.5) {$>>$};

    \draw[->] (2, 1) to[bend left] (5,1);

    \draw (4, -0.5) arc(-90:90:0.25 and 0.5)
      node[midway,rotate=90] {$>>$};
    \draw (4, -0.5) arc(-90:-270:0.25 and 0.5);
    \draw (4, 0.5) -- (6, 0.5) node[midway] {$>$};
    \draw (4, -0.5) -- (6, -0.5) node[midway] {$>$};
    \draw (6, -0.5) arc(-90:90:0.25 and 0.5)
      node[midway,rotate=90] {$>>$};
    \draw[dashed] (6, -0.5) arc(-90:-270:0.25 and 0.5);

    \node at (8, 0) {$\simeq \mathbb{R}^2 / G$};
  \end{tikzpicture}
  \caption{Klein's figure on a Cartesian plane to a cylinder}
  \label{fig:klein_s_figure_on_a_cartesian_plane_to_a_cylinder}
\end{figure*}

\noindent
where we define $G = \left\{ T^k \mmid k \in \mathbb{Z} \right\}$ as before,
for $T ((x, y)) = (x + 1, y)$. Now on a similar note, define
$R : \mathbb{R} \to \mathbb{R}$ by $R((x, y)) = (x, y + 1)$. Then let
$G_2 = \left\{ R^k \mmid k \in \mathbb{Z} \right\}$. Thus
\begin{equation*}
  \mathbb{R}^2 / G \oplus G_2 \simeq \mathbb{R}^2 / \mathbb{Z} \oplus \mathbb{Z}.
\end{equation*}

% subsubsection the_space_of_s_1_times_s_1_ (end)

% subsection the_1_sphere_s_1_continued (end)

% section manifolds_continued_2 (end)

% chapter lecture_6_jan_18th (end)

\chapter{Lecture 7 Jan 21st}%
\label{chp:lecture_7_jan_21st}
% chapter lecture_7_jan_21st

\section{Manifolds (Continued 3)}%
\label{sec:manifolds_continued_3}
% section manifolds_continued_3

\subsection{\texorpdfstring{$2$}{2}-dimensional Manifolds}%
\label{sub:_2_dimensional_manifolds}
% subsection _2_dimensional_manifolds

\begin{eg}\label{eg:2_sphere_as_a_manifold}
  $S^2$ is a $2$-dimensional manifold (w/o boundary).
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \pgfmathsetmacro\rad{1};
      \draw (0, 0, 0) circle(\rad);
      \shade[ball color=blue!70!foreground,opacity=0.5] (0, 0, 0) circle(\rad);
      \draw[opacity=0.5] (-\rad,0,0) arc (180:360:{\rad} and {\rad/2});
      \draw[dashed,opacity=0.5] (-\rad,0,0) arc (180:0:{\rad} and {\rad/2});
      \draw[opacity=0.5] (0,\rad,0) arc (90:270:{\rad/2} and \rad);
      \draw[dashed,opacity=0.5] (0,\rad,0) arc (90:-90:{\rad/2} and \rad);

      \fill[color=red,opacity=0.3] (-{\rad}, 0, 0) arc (180:360:{\rad}) -- ({\rad}, 0, 0) arc (360:180:{\rad} and {\rad/2});
      \fill[color=red,opacity=0.2] (-{\rad}, 0, 0) arc (180:0:{\rad} and {\rad/2}) -- ({\rad}, 0, 0) arc (360:180:{\rad} and {\rad/2});

      % the axes
      \draw[dashed] (-2, 0, 0) -- (0, 0, 0);
      \draw[dashed] (0, -2, 0) -- (0, 0, 0);
      \draw[dashed] (0, 0, -2) -- (0, 0, 0);
      \draw[-latex'] (0, 0, 0) -- (2, 0, 0) node[right] {$y$};
      \draw[-latex'] (0, 0, 0) -- (0, 2, 0) node[above] {$z$};
      \draw[-latex'] (0, 0, 0) -- (0, 0, 2) node[below,left] {$x$};
    \end{tikzpicture}
    \caption{The $2$-sphere $S^2$}
    \label{fig:the_2_sphere_s_2_}
  \end{marginfigure}
  Note that we may `cover' this sphere by the function $f_{-} : \mathbb{D} \subseteq \mathbb{R}^2 \to \mathbb{R}$,
  where $\mathbb{D}$ is the unit disc in $\mathbb{R}^2$, and $f_{-}$ is given by
  \begin{equation*}
    (x, y) \mapsto - \sqrt{1 - (x^2 + y^2)}.
  \end{equation*}
  We see that $f_{-}$ is a chart for the lower hemisphere for \cref{fig:the_2_sphere_s_2_}, shaded red.
  We can indeed cover the entire $2$-sphere with similar hemispheres in different orientations: consider
  \begin{gather*}
    f_+ (x, y) = \sqrt{ 1 - x^2 - y^2 } \\
    g_+ (x, z) = \sqrt{ 1 - x^2 - z^2 } \\
    g_- (x, z) = - \sqrt{ 1 - x^2 - z^2 } \\
    h_+ (y, z) = \sqrt{ 1 - y^2 - z^2 } \\
    h_- (y, z) = - \sqrt{ 1 - y^2 - z^2 },
  \end{gather*}
  which represent the upper hemisphere, eastern hemisphere, western hemisphere, front hemisphere,
  and back hemisphere, respectively.
\end{eg}

We have the following general observation: let $f : \mathcal{U} \subseteq \mathbb{R}^n \to \mathbb{R}$
be a real-valued continuous function. Then $\Gamma(f)$ is \hlnotea{parameterized} using $f$, i.e. $f$
constructs the chart $\Gamma(f)$.

We have not explicitly defined what $\Gamma(f)$ is, but it is recorded
in \href{https://tex.japorized.ink/PMATH733/classnotes.pdf\#page.32}{PMATH 733}, but we shall provide a
short definition here for reference: the \hldefn{graph} $\Gamma(f)$ of the function $f$ is defined as
\begin{equation*}
  \Gamma(f) := \left\{ (x_1, \ldots, x_n, y) \mmid f(x_1, \ldots, x_n) = y \right\}.
\end{equation*}

It is interesting to note that $\Gamma(f)$ is homeomorphic to $\mathcal{U}$. Consider the function
\begin{equation*}
  \Phi : \mathcal{U} \to \Gamma(F) \text{ given by } \Phi(x_1, \ldots, x_n) = (x_1, \ldots, x_n, f(x_1, \ldots, x_n)).
\end{equation*}
It is clear that $\Phi$ is continuous since each of its components are continuous. However, it is not
as easy to find a continuous map to go from $\Gamma(f)$ to $\mathcal{U}$.

\begin{defn}[Projection]\index{Projection}\label{defn:projection}
  We define function $\pi_{n + 1} : \Gamma(f) \to \mathcal{U}$ by
  \begin{equation*}
    \pi_{n + 1}(x_1, \ldots, x_n, y) = (x_1, \ldots, x_n),
  \end{equation*}
  which is called a \hlnoteb{projection}.
\end{defn}

\begin{note}
  \begin{enumerate}
    \item In A1, we showed that $\pi_{n + 1}$ is continuous.
    \item Furthermore, $\pi_{n + 1}$ is injective.
    \item Thus, we observe that
      \begin{equation*}
        \pi_{n + 1} \circ \Phi(x_1, \ldots, x_n) = (x_1, \ldots, x_n) = id_{n + 1}(x_1, \ldots, x_n).
      \end{equation*}
  \end{enumerate}
\end{note}

\begin{eg}
  Consider the following figure:
  \begin{figure}[ht]
    \centering
    \begin{tikzpicture}
      \draw (0, 0, 0) circle(2);
      \shade[ball color=blue!70!foreground,opacity=0.5] (0, 0, 0) circle(2);
      \draw[opacity=0.5] (-2,0,0) arc (180:360:2 and 1);
      \draw[dashed,opacity=0.5] (-2,0,0) arc (180:0:2 and 1);
      \draw[opacity=0.5] (0,2,0) arc (90:270:1 and 2);
      \draw[dashed,opacity=0.5] (0,2,0) arc (90:-90:1 and 2);
      \draw (0, 0, 0) circle(2);

      % Stereographic projection
      \node[draw,circle,color=red,fill,inner sep=1.5pt,pin={[pin edge=latex'-, pin distance=5pt]45:{$N$, the north pole}}] at (0, 2, 0) {};
      \node[draw,circle,color=red,fill,inner sep=1.5pt,pin={[pin edge={latex'-}, pin distance=40pt]30:{$p$}}] at (1.5, 1, 0.5) {};
      \node[draw,circle,color=red,fill,inner sep=1.5pt,pin={[pin edge=latex'-, pin distance=20pt]245:{$p'$}}] at (3, 0, 1) {};
      \draw[thick,red] (0, 2, 0) -- (1.5,1,0.5) -- (3, 0, 1);

      \draw[dashed] (3, 0, 1) -- (0, 0, 0);
      \draw[dashed] (3, 0, 0) -- (3, 0, 1) -- (0, 0, 1);
      \draw[dashed] (0, 0, 0) -- (1.5, 1, 0.5) -- (1.5, 0, 0.5) -- (0, 0, 0.5);
      \draw[dashed] (1.5, 0, 0) -- (1.5, 0, 0.5);

      % the axes
      \draw[dashed] (-3, 0, 0) -- (0, 0, 0);
      \draw[dashed] (0, -3, 0) -- (0, 0, 0);
      \draw[dashed] (0, 0, -3) -- (0, 0, 0);
      \draw[-latex'] (0, 0, 0) -- (3.5, 0, 0) node[right] {$y$};
      \draw[-latex'] (0, 0, 0) -- (0, 3, 0) node[above] {$z$};
      \draw[-latex'] (0, 0, 0) -- (0, 0, 3) node[below,left] {$x$};
    \end{tikzpicture}
    \caption{Stereographic Projection}
    \label{fig:stereographic_projection}
  \end{figure}

  We define the \hldefn{stereographic projection} by
  \begin{equation*}
    \Sigma(p) = p'.
  \end{equation*}
\end{eg}

% subsection _2_dimensional_manifolds (end)

\subsection{\texorpdfstring{$n$}{n}-spheres}%
\label{sub:_n_spheres}
% subsection _n_spheres

We can now extend the notion we saw in \cref{eg:2_sphere_as_a_manifold} to higher dimensional
manifolds. In particular, we want to be able to `glue' the boundaries of the hemispheres, as
shown in \cref{fig:glueing_two_hemispheres}.

\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \draw[dashed] (-2, 1, 0) -- (-2, -1.5, 0);
    \draw[dashed] (2, 1, 0) -- (2, -1.5, 0);
    \draw[dashed] (-1.5, 1, -2.3) -- (-1.5, -1.5, -2.3);
    \draw[dashed] (0, 1, -2.3) -- (0, -1.5, -2.3);
    \draw[dashed] (1, 1, 2.6) -- (1, -1.5, 2.6);

    \draw[dashed] (-2, 1, 0) arc (180:0:2 and 1);
    \draw[fill=red,fill opacity=0.5] (-2, 1, 0) arc (180:360:2 and 1) -- (2, 1, 0) arc (0:180:2);

    \draw[fill=background,fill opacity=0.5] (-2, -1.5, 0) arc (180:360:2) -- (2, -1.5, 0) arc (0:180:2 and 1) -- (-2, -1.5, 0) arc (180:360:2 and 1);

    \begin{scope}
      \clip (-2, 1, 0) arc (180:360:2 and 1) -- (2, 1, 0) arc (0:180:2);
      \draw[fill=foreground!70!background] (1, 1, 2.6) circle(0.5);
    \end{scope}
    \node[draw,circle,fill] at (1, 1, 2.6) {};

    \begin{scope}
      \clip (-2, -1.5, 0) arc (180:360:2) -- (2, -1.5, 0) arc (360:180:2 and 1);
      \draw[fill=foreground!70!background] (1, -1.5, 2.6) circle(0.5);
    \end{scope}
    \node[draw,circle,fill] at (1, -1.5, 2.6) {};
  \end{tikzpicture}
  \caption{Glueing two hemispheres}\label{fig:glueing_two_hemispheres}
\end{marginfigure}
\marginnote{I was definitely not trying to make a Pokeball.}

In a more mathematical sense, we are identifying the lower and upper boundaries of the upper
and lower hemisphere, respectively, i.e. we identify
\begin{equation*}
  S^n = \bar{\mathbb{B}}^n \times \{ 0 \} \cup \bar{\mathbb{B}}^n \times \{ 1 \},
\end{equation*}
where the $\{ 0 \}$ and $\{ 1 \}$ represent the upper and lower hemispheres, respectively. We
may also write this as
\begin{equation*}
  x \in \partial \bar{\mathbb{B}}^n : (x, 0) \sim (x, 1).
\end{equation*}

This calls for the notion of an equivalence class. Recall that an \hldefn{equivalence relation}
on a set $A$ is a relation between elements such that
\begin{enumerate}
  \item (reflexity) $x \sim x$;
  \item (symmetry) $x \sim y \iff y \sim x$; and
  \item (transitivity) $x \sim y \, \land \, y \sim z \implies x \sim z$,
\end{enumerate}
where $\sim$ is our equivalence relation.

Equivalence relations give rise to the notion of \hldefn{equivalence classes}, where we define
an equivalence class as
\begin{equation*}
  [\beta] := \{ \alpha \in A \mid \beta \sim \alpha \}.
\end{equation*}
In this course, we shall sometimes denote equivalence classes in the form of $A_\beta$. Notice
that
\begin{equation*}
  A = \coprod\limits_{\beta \in B} A_\beta,
\end{equation*}
where $B \subseteq A$.

\newthought{This begs the question}: do the set of equivalence classes retain the topology of
the space?

Here, we look into \hldefn{quotient topology}. Let $A^*$ be the space of equivalence classes,
and let's assume that $A$ is endowed with a topology. Consider the function
\begin{equation*}
  \pi : A \to A^* \text{ given by } a \mapsto [a]_{\sim},
\end{equation*}
where $\sim$ is the equivalence relation, and $[a]_{\sim}$ is the equivalence class that $a$
belongs to\sidenote{Note that the function is well-defined as the equivalence classes are
disjoint.}.

\begin{figure}[ht]
  \centering
  \begin{tikzcd}
    A \arrow[dd, "\pi"'] \arrow[rrdd, "g"] &  &  \\
     &  &  \\
    A^* \arrow[rr, "\tilde{g}"'] &  & Z
  \end{tikzcd}
  \caption{Relationship of $A$, $A^*$ and $Z$}
  \label{fig:relationship_of_a_a_and_z_}
\end{figure}

Consider an arbitrary space $Z$ that is also endowed with a topology, such that
$\exists g : A \to Z$ that is continuous, such that $g$ is constant on each of the equivalence
classes of $A$, that is, if $a \sim b$, then $g(a) = g(b)$.

Then $\tilde{g}$ is a well-defined function induced on $A^*$: we have that
\begin{equation*}
  \tilde{g}(\pi(a)) = g(a).
\end{equation*}

We want to endow $A^*$ with a topology such that $\tilde{g}$ will be continuous. However,
\begin{itemize}
  \item if $A^*$ is too `fine', then $\pi$ may not be continuous; and
  \item if $A^*$ is too `course', then $\tilde{g}$ may not be continuous.
\end{itemize}
We need to strike a balance in the fineness of the topology of $A^*$ to make sure that both
$\pi$ and $\tilde{g}$ are continuous.

\begin{defn}[Strongly Continuous]\index{Strongly Continuous}\label{defn:strongly_continuous}
  Let $\mathcal{V} \subseteq A^*$. $\mathcal{V}$ is open iff $\exists \mathcal{U} \subseteq A$
  such that $\pi(\mathcal{U}) = \mathcal{V}$. We say that $\pi$ is \hlnoteb{strongly continuous}.
\end{defn}

\begin{warning}
  $\pi$ may not be an open map!
\end{warning}

% subsection _n_spheres (end)

% section manifolds_continued_3 (end)

% chapter lecture_7_jan_21st (end)

\chapter{Lecture 8 Jan 23rd}%
\label{chp:lecture_8_jan_23rd}
% chapter lecture_8_jan_23rd

\section{Quotient Spaces}%
\label{sec:quotient_spaces}
% section quotient_spaces

With the topology that we last introduced for $A^*$, $g$ is continuous iff $\tilde{f}$ is continuous.
\marginnote{This way of constructing a quotient space is intuitive, ring theoretically.}
However, this is a rather cumbersome way to construct a quotient space. In particular, how do we know
what equivalence class should we choose?

\marginnote{
\begin{ex}
  Let $\pi : X \to Y$ be any map. For a subset $U \subseteq X$, show that TFAE.
  \begin{enumerate}
    \item $U$ is saturated.
    \item $U = \pi^{-1}(q(U))$.
    \item $U$ is a union of fibres.
    \item If $x \in U$, then every point $x' \in X$ such that $q(x) = q(x')$ is also in $U$.
  \end{enumerate}
\end{ex}
}
\begin{defn}[Saturated]\index{Saturated}\label{defn:saturated}
  We say that a subset $S \subseteq A$ is \hlnoteb{saturated} (wrt $\pi$) provided that it is has
  a \hlimpo{non-empty intersection} with a fibre $\pi^{-1}(\{ q \})$, where $q \in A^*$, then
  $S \supseteq \pi^{-1}(\{ q \})$.
\end{defn}

\begin{note}
  This definition is equivalent to saying that $\pi^{-1} \circ \pi (S) = S$.
\end{note}

With this definition, we may restate the definition of a quotient map in a clearer way.

\begin{defn}[Quotient Map]\index{Quotient Map}\label{defn:quotient_map}
  A map $g : X \to Y$ is called a \hlnoteb{quotient map} if it sends saturated open subsets of $X$
  to open subsets of $Y$.
\end{defn}

\begin{eg}
  Let $g : X \to Y$ be surjective. We can define an equivalence relation on $X$ by setting
  $x_1 \sim x_2$ iff $g(x_1) = g(x_2)$, i.e. $x_1$ and $x_2$ belong to the same fibre.
\end{eg}

\begin{eg}
  Let $\bar{\mathbb{B}}^2$ be the closed unit disk in $\mathbb{R}^2$, and let $\sim$ be the
  equivalence relation on $\bar{\mathbb{B}}^2$ defined by $(x, y) \sim (-x, y)$ for all $(x, y)
  \in \partial \bar{\mathbb{B}}^2$ (See \cref{fig:a_quotient_of_b_2}).
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw[fill=blue!30!background] (0, 0) circle(1);
      \draw[midarrow] (0, 1) arc(90:270:1);
      \draw[midarrow] (0, 1) arc(90:-90:1);
      \node[draw,circle,fill,inner sep=1pt] at (0, 1) {};
      \node[draw,circle,fill,inner sep=1pt] at (0, -1) {};
    \end{tikzpicture}
    \caption{A quotient of $\bar{\mathbb{B}}^2$}\label{fig:a_quotient_of_b_2}
  \end{marginfigure}
  We can think of this space as one that is obtained from $\mathbb{B}^2$ by ``pasting'' the
  left half of the boundary to the right half. It is not difficult to imagine this transformation
  and notice that we can `continuously morph' $\mathbb{B}^2$ under this equivalence relation into
  $S^2$. We shall prove much later on that this is indeed the case.

  The above process is also called `collapsing $\partial \mathbb{B}^2$ to a point'.
\end{eg}

\begin{eg}
  Define an equivalence relation on the square $I \times I$ by setting $(x, 0) \sim (x, 1)$ for
  all $x \in I$, and $(0, y) \sim (1, y)$ for all $y \in I$ (See
  \cref{fig:a_quotient_of_i_times_i_}).
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw[fill=blue!30!background] (-1, 1) -- (-1, -1) -- (1, -1) -- (1, 1) --cycle;
      \draw[midarrow={>>}{0.5}] (-1, 1) -- (-1, -1);
      \draw[midarrow={>>}{0.5}] (1, 1) -- (1, -1);
      \draw[midarrow={>}{0.5}] (1, 1) -- (-1, 1);
      \draw[midarrow={>}{0.5}] (1, -1) -- (-1, -1);
    \end{tikzpicture}
    \caption{A quotient of $I \times I$}\label{fig:a_quotient_of_i_times_i_}
  \end{marginfigure}
\end{eg}

\begin{eg}\label{eg:projective_space}
  Define $\mathbb{P}^n$, the \hldefn{real projective space of dimension $n$}, to be the set of
  $1$-dimensional linear subpaces (lines through the origin) in $\mathbb{R}^{n + 1}$. There
  exists a map $q : \mathbb{R}^{n + 1} \setminus \{ 0 \} \to \mathbb{P}^n$, defined by sending
  a point $x$ to its span. We apply the quotient topology with respect to $q$ on $\mathbb{P}^2$.
  The $2$-dimensional projective space $\mathbb{P}^2$ is usually called the \hldefn{projective
  plane}.
\end{eg}

\begin{eg}
  Let $X$ be a topological space, the quotient $(X \times I) / (X \times \{ 0 \})$ obtained from
  the ``cylinder'' $X \times I$ by collapsing one end to a point is called the \hldefn{cone} on
  $X$. For instance, if $X = S^1$ and $I = (0, 1)$, then taking the quotient
  $(S^1 \times (0, 1)/(S^1 \times \{ 0 \})$ is exactly the process shown in
  \cref{fig:cone_of_s_1_}.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (-1, 1) -- (-1, -2) arc(180:360:1 and 0.5) -- (1, -2) --
        (1, 1) arc(0:180:1 and 0.5) -- (-1,1) arc(180:360:1 and 0.5);
      \draw[dashed] (-1, -2) arc(180:0:1 and 0.5);

      \node[rotate=-90,scale=1.5] at (0, -3) {$\leadsto$};

      \draw (0, -4) -- (-1, -7) arc(180:360:1 and 0.5) -- (1, -7) -- cycle;
      \draw[dashed] (-1, -7) arc(180:0:1 and 0.5);
    \end{tikzpicture}
    \caption{Cone of $S^1$}\label{fig:cone_of_s_1_}
  \end{marginfigure}
\end{eg}

\begin{eg}[Wedge sum of Manifolds]\label{eg:wedge_sum_of_manifolds}
  Let $M_1$ and $M_2$ be two manifolds. We define the \hldefn{wedge sum} of $M_1$ and $M_2$ as
  \begin{equation*}
    M_1 \lor M_2 := (M_1 \cup M_2) / p_1 \sim p_2,
  \end{equation*}
  where $p_1 \in M_1$ and $p_2 \in M_2$. The wedge sum is also sometimes called the
  \hldefn{one-point union}.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw[latex-latex] (-1, 0) -- (1, 0);
      \draw[latex-latex] (0, -1) -- (0, 1);
      \node[draw,circle,fill,inner sep=1.5pt] at (0, 0) {};
    \end{tikzpicture}
    \caption{Wedge sum of two lines.}\label{fig:wedge_sum_of_two_lines}
  \end{marginfigure}
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (0, 1) circle(1);
      \draw (0, -1) circle(1);
      \node[draw,circle,fill,inner sep=1.5pt] at (0, 0) {};
    \end{tikzpicture}
    \caption{Wedge sum of two circles.}\label{fig:wedge_sum_of_two_circles}
  \end{marginfigure}
  For instance, the wedge sum of $\mathbb{R} \wedge \mathbb{R}$ is homeomorphic to the union
  of the $x$-axis and $y$-axis on a Cartesian plane (cf \cref{fig:wedge_sum_of_two_lines}),
  and the wedge sum $S^1 \wedge S^1$ is homeomorphic to the \hldefn{figure-eight space},
  which is made up by the union of two circles of radius $1$ centered at $(0, 1)$ and $(0, -1)$
  in the plane (cf \cref{fig:wedge_sum_of_two_circles}).
\end{eg}

\begin{warning}
  Unlike subspaces and product spaces, quotient spaces do not behave well wrt most topological
  properties. In particular, none of the \hyperref[defn:manifold]{definiting properties of
  manifolds} are necessarily inherited by quotient spaces.

  % TODO : Provide examples
  For example, a quotient space can be \sidenote{\hlwarn{I should provide
  examples here}.}
  \begin{itemize}
    \item locally Euclidean and second countable, but not Hausdorff;
    \item Hausdorff and second countable, but not locally Euclidean; or
    \item not even first countable.
  \end{itemize}
\end{warning}

The following is a proposition that gives us some peace of mind when working within certain
spaces.

\begin{propo}[Locally Euclidean Quotient Space of a Second Countable Space is Second Countable]\label{propo:locally_euclidean_quotient_space_of_a_second_countable_space_is_second_countable}
  Suppose $M$ is a second countable space and $N$ is a quotient space of $M$. If $N$ is
  locally Euclidean, then it is second countable.
\end{propo}

\begin{note}
  Thus if the original space is, say, a manifold, then for any of its quotient spaces, we only
  need to check that the quotient space is both Hausdorff and locally Euclidean.
\end{note}

\begin{proof}
  Let $q : M \to N$ be the quotient map, and suppose $N$ is locally Euclidean. Let $\mathcal{U}$
  be a cover of $N$. Then the set $\left\{ q^{-1}(U) : U \in \mathcal{U} \right\}$ is an open
  cover of $M$, which therefore has a countable subcover. Let $\mathcal{U}' \subseteq \mathcal{U}$
  be the countable subset such that $\left\{ q^{-1}(U) : U \in \mathcal{U}' \right\}$ covers $M$.
  Then $\mathcal{U}'$ is a countable subcover of $N$.
\end{proof}

\begin{ex}
  Consider the function $g : \mathbb{C} \to \mathbb{C}$ given by $z \mapsto z^2$. Verify that
  $g$ is a quotient map.
\end{ex}

\begin{eg}
  The map $g : S^1 \to S^1 \subseteq \mathbb{R}^2$ as given by the above is indeed a quotient map.
  Thus we observe that $S^1$ is a quotient space of itself.
\end{eg}

\subsection{Characteristic Property and Uniqueness of Quotient Spaces}%
\label{sub:characteristic_property_and_uniqueness_of_quotient_spaces}
% subsection characteristic_property_and_uniqueness_of_quotient_spaces

\begin{thm}[Characteristic Property of the Quotient Topology]\label{thm:characteristic_property_of_the_quotient_topology}
  Suppose $X$ and $Y$ are two topological spaces and $\pi : X \to Y$ is a quotient map. For any
  topological space $Z$, a map $f : Y \to Z$ is continuous iff the composite map $f \circ \pi$
  is continuous (cf \cref{fig:characteristic_property_of_the_quotient_topology_}).
  \begin{marginfigure}
    \centering
    \begin{tikzcd}
      X \arrow[rrdd, "f \circ \pi"] \arrow[dd, "\pi"'] &  &  \\
       &  &  \\
      Y \arrow[rr, "f"'] &  & Z
    \end{tikzcd}
    \caption{Characteristic property of the quotient topology.}\label{fig:characteristic_property_of_the_quotient_topology_}
  \end{marginfigure}
\end{thm}

\begin{proof}
  Observe that for any open $U \subseteq Z$, $f^{-1}(U)$ is open in $Y$ iff
  \begin{equation*}
    \pi^{-1}(f^{-1}(U)) = (f \circ \pi)^{-1}(U)
  \end{equation*}
  is open in $X$. Our result follows immediately from this observation.
\end{proof}

\begin{thm}[Uniqueness of the Quotient Topology]\label{thm:uniqueness_of_the_quotient_topology}
  Given a topological space $X$, a set $Y$ and a surjective map $\pi : X \to Y$, the quotient
  topology is the only topology on $Y$ for which the characterisitc property holds.
\end{thm}

\marginnote{
\begin{ex}
  Prove \cref{thm:uniqueness_of_the_quotient_topology}.
\end{ex}
}

\begin{thm}[Descends to the Quotient]\label{thm:descends_to_the_quotient}
  Suppose $\pi : X \to Y$ is a quotient map, $Z$ a topological space, and $f : X \to Z$ is any
  continuous map that is constant on the fibres of $\pi$ \sidenote{This means that if $\pi(x)
  = \pi(x')$, then $f(x) = f(x')$.}. Then there exists a unique continuous map $\tilde{f} : Y
  \to Z$ such that $f = \tilde{f} \circ \pi$ (cf \cref{fig:descends_to_the_quotient}).
  \begin{marginfigure}
    \centering
    \begin{tikzcd}
      X \arrow[rrdd, "f"] \arrow[dd, "\pi"'] &  &  \\
       &  &  \\
      Y \arrow[rr, "\tilde{f}"', dashed] &  & Z
    \end{tikzcd}
    \caption{Descends to the Quotient}\label{fig:descends_to_the_quotient}
  \end{marginfigure}
\end{thm}

\begin{proof}
  Since $\pi$ is surjective, $\forall y \in Y, \, \exists x \in X$ such that $\pi(x) = y$. Then
  consider $\tilde{f} : Y \to Z$ given by $\tilde{f}(y) = f(x)$ for any $x$ that we discover from
  the last statement. By the hypothesis on $f$, $\tilde{f}$ is guaranteed to be unique and
  well-defined. It follows from \cref{thm:characteristic_property_of_the_quotient_topology} that
  $\tilde{f}$ is continuous.
\end{proof}

The following theorem, which is a consequence of
\cref{thm:characteristic_property_of_the_quotient_topology}, gives to us that quotient spaces are
unique up to homeomorphism by the identifications made by their quotient maps.

\begin{thm}[Uniqueness of Quotient Spaces]\label{thm:uniqueness_of_quotient_spaces}
  Suppose $\pi_1 : X \to Y_1$ and $\pi_2 : X \to Y_2$ are quotient maps that make the same
  identifications, i.e.
  \begin{equation*}
    \pi_1(x) = \pi_1(x') \iff \pi_2(x) = \pi_2(x').
  \end{equation*}
  Then there exists a unique homeomorphism $\phi : Y_1 \to Y_2$ such that $\pi \circ \pi_1 =
  \pi_2$.
\end{thm}

The proof is relatively straightforward so I will only jot down the gist of the proof and provide
commutative diagrams for visualization of the relationships between these spaces.

\begin{proof}[Sketch]
  Observe \cref{fig:relationships_of_the_quotient_spaces}.
  \begin{figure}[ht]
    \centering
    \begin{tikzcd}
      X \arrow[rrdd, "\pi_2"] \arrow[dd, "\pi_1"'] &  &  & X \arrow[dd, "\pi_2"'] \arrow[rrdd, "\pi_1"] &  &  \\
       &  &  &  &  &  \\
      Y_1 \arrow[rr, "\tilde{\pi}_2"', dashed] &  & Y_2 & Y_2 \arrow[rr, "\tilde{\pi}_1"', dashed] &  & Y_1
    \end{tikzcd}
    \caption{Relationships of the quotient spaces}
    \label{fig:relationships_of_the_quotient_spaces}
  \end{figure}

  Then
  \begin{equation*}
    \tilde{\pi}_1 \circ (\tilde{\pi}_2 \circ \pi_1) = \tilde{\pi}_1 \circ \pi_2 = \pi_1.
  \end{equation*}
  Thus, we have \cref{fig:consequence_of_the_relationship_between_the_quotient_spaces_}.
  \begin{figure}[ht]
    \centering
    \begin{tikzcd}
      X \arrow[rrdd, "\pi_1"] \arrow[dd, "\pi_1"'] &  &  \\
       &  &  \\
      Y_1 \arrow[rr, "\tilde{\pi}_1 \circ \tilde{\pi}_2"', dashed] &  & Y_1
    \end{tikzcd}
    \caption{Consequence of the relationship between the quotient spaces.}
    \label{fig:consequence_of_the_relationship_between_the_quotient_spaces_}
  \end{figure}

  Then by \cref{thm:uniqueness_of_the_quotient_topology}, the map is indeed unique, and it
  follows from the identity map that the two maps are equal. Similarly, $\tilde{\pi}_2 \circ
  \tilde{\pi}_1$ is the identity on $Y_2$.

  Then $\phi = \tilde{\pi}_2$ is the required homeomorphism, and it is unique by
  \cref{thm:uniqueness_of_the_quotient_topology}.

\end{proof}

% subsection characteristic_property_and_uniqueness_of_quotient_spaces (end)

% section quotient_spaces (end)

% chapter lecture_8_jan_23rd (end)

\chapter{Lecture 9 Jan 25th}%
\label{chp:lecture_9_jan_25th}
% chapter lecture_9_jan_25th

\section{Topological Embeddings}%
\label{sec:topological_embeddings}
% section topological_embeddings

\begin{defn}[Topological Embedding]\index{Topological Embedding}\label{defn:topological_embedding}
  An injective continuous map $g : S \to Y$ is called a \hlnoteb{topological embedding} (or just
  an \hldefn{embedding}) if it is a homeomorphism onto its image.
\end{defn}

\begin{note}
  In other words, $g : S \to Y$ is called an embedding if it is a homeomorphism between $S$ and
  $g(S)$.
\end{note}

\begin{eg}
  Consider the function $f : \mathbb{R} \to \mathbb{R}^3$ given by
  \begin{equation*}
    x \mapsto (\cos x, \sin x, x).
  \end{equation*}
  We know that $f$ is both bijective and continuous (since each of its components are continuous).
  Note that
  \begin{equation*}
    f \circ \pi_3 \restriction_{f(\mathbb{R})} = \id_{\mathbb{R}}.
  \end{equation*}
  It follows that $f$ is an embedding.
\end{eg}

\begin{eg}
  The map $f : x \mapsto e^{ix}$ from $[0, 2 \pi)$ to $\mathbb{C}$ is continuous and injective, but
  not a homeomorphism (problem lies on the endpoints). So $f$ is not an embedding, despite being
  continuous and injective.

  However, the restriction of $f$ to any proper subinterval is an embedding, as is the interval
  $(0, 2 \pi)$.
\end{eg}

As we've seen above, a continuous injective map is not necessarily an embedding. The following
proposition provides us with two sufficient but not necessary conditions to ensure that a
continuous injective map is an embedding.

\marginnote{
\begin{marginwarning}
  Notice that by our definition, and by taking note on \cref{ex:homeomorphm_iff_open_and_closed},
  an embedding is \hlimpo{only a homeomorphism between the domain and its image} under the mapping.
  It need not be a homeomorphism between the domain and the codomain.  Thus, an embedding need not
  be open or closed.
\end{marginwarning}

\begin{eg}
  The map $f : \left(0, \frac{1}{2}\right) \to S^1$ given by $x \to e^{2 \pi i x}$ is an
  embedding but it is neither open nor closed.
\end{eg}

\begin{ex}
  Give another example of a topological embedding that is neither open nor closed.
\end{ex}
}

\begin{propo}[Sufficient Conditions to be an Embedding]\label{propo:sufficient_conditions_to_be_an_embedding}
  A continuous injective map that is either open or closed is an embedding.
\end{propo}

\begin{proof}
  Let $f : X \to Y$ be a continuous injective map between 2 topological spaces. Note that $f$
  is bijective from $X$ to $f(X)$. It suffices to show that $f$ is open or closed from $X$ to
  $f(X)$ by \cref{ex:homeomorphm_iff_open_and_closed}.

  By assumption, if $f : X \to Y$ is open, then for any open $A \subset X$, $f(A) \subset f(X)$
  is open, simply by \hyperref[defn:open_and_closed_maps]{definition}. It also follows that if
  $f$ is closed, then for any closed $B \subset X$, $f(B)$ is closed in $f(X)$.
\end{proof}

\marginnote{
\begin{ex}
  Prove \cref{propo:surjective_embeddings_are_homeomorphisms}.
\end{ex}
}
\begin{propo}[Surjective Embeddings are Homeomorphisms]\label{propo:surjective_embeddings_are_homeomorphisms}
  A surjective topological embedding is a homeomorphism.
\end{propo}

\begin{remark}\label{remark:we_are_ready_to_make_more_manifolds}
  We now have some proper tools to build various examples of manifolds as subspaces of Euclidean
  spaces. Recall \cref{propo:other_properties_of_the_subspace_topology} and
  \cref{propo:subspaces_of_second_countable_spaces_are_second_countable}, to show that a subspace
  of $\mathbb{R}^n$ is a manifold, we need only to verify that it is locally Euclidean.
\end{remark}

\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \begin{axis}[tufteaxis, range3frame, view={55}{45}]
      \addplot3[surf, colormap/hot, samples=5, opacity=0, domain=40:140] {2};
      \addplot3[surf, colormap/hot, samples=10, domain=60:120] {0};
      \addplot3[surf, colormap/hot2, samples=20, domain=60:120] {sin(4*x)*sin(4*y) + 2};
      \draw[dashed] (60, 60, 2.7) -- (60, 60, 0);
      \draw[dashed] (120, 120, 2.7) -- (120, 120, 0);
      \draw[dashed] (120,60,1.2) -- (120,60,0);
      \node[pin={[pin distance=5pt]135:{$U$}}] at (60, 80, 0) {};
      \node[pin={[pin distance=7pt]135:{$\Gamma(f)$}}] at (70, 70, 3) {};
    \end{axis}
  \end{tikzpicture}
  \caption{Graph of a continuous function}\label{fig:graph_of_a_continuous_function}
\end{figure*}

\begin{eg}
  If $U \subseteq \mathbb{R}^n$ is an open subset and $f : U \to \mathbb{R}^k$ is any continuous
  map, the graph of $f$ (cf \cref{fig:graph_of_a_continuous_function}) is the subset $\Gamma(f)
  \subseteq \mathbb{R}^{n + k}$ defined by
  \begin{equation*}
    \Gamma(f) = \{(x, y) = (x_1, \ldots, x_n, y_1, \ldots, y_k) : x \in U, \, y = f(x)\},
  \end{equation*}
  with the subspace topology inherited from $\mathbb{R}^{n + k}$. To verify that $\Gamma(f)$ is
  indeed a manifold, we need only to show that $\Gamma(f)$ is homeomorphic to $U$. Let $\phi_f
  : U \to \mathbb{R}^{n + k}$ be the continuous injective map
  \begin{equation*}
    \phi_f(x) = (x, f(x)).
  \end{equation*}
  One can verify that $\phi_f$ is indeed a continuous bijection from $U$ to $\Gamma(f)$, and for
  $\pi : \mathbb{R}^{n + k} \to \mathbb{R}^n$, $\pi_{\Gamma(f)}$ is a continuous inverse for
  $\phi_f$. It follows that $\phi_f$ is an embedding and $\Gamma(f)$ is homeomorphic to $U$.
\end{eg}

\begin{eg}[$n$-spheres are Manifolds]\label{eg:n_spheres_are_manifolds}
  Recall that the $n$-sphere (or \textit{unit}) $n$-sphere is the set $S^n$ of unit vectors in
  $\mathbb{R}^n$. In low dimensions, spheres are easy to visualize:
  \begin{itemize}
    \item $S^0$ is the two-point discrete space $\{ \pm 1 \} \subseteq \mathbb{R}$;
    \item $S^1$ is the unit circle in $\mathbb{R}^2$; and
    \item $S^2$ is the unit spherical surface of radius $1$ in $\mathbb{R}^3$.
  \end{itemize}
  Since we are working in $\mathbb{R}^n$, by \cref{remark:we_are_ready_to_make_more_manifolds},
  it suffices for us to show that each of the $S^n$'s are locally Euclidean. We shall show that
  each point has a neighbourhood in $S^n$ that is the graph of a continuous function.

  For each $i \in \{1, \ldots, n + 1 \}$, let
  \begin{itemize}
    \item[$U_i^+$] denote the open subset of $\mathbb{R}^{n + 1}$ consisting of points with
      $x_i > 0$, and
    \item[$U_i^-$] denote the open subset of $\mathbb{R}^{n + 1}$ consisting of points with
      $x_i < 0$.
  \end{itemize}
  Then for any $x = (x_1, \ldots, x_n) \in S^n$, some coordinate $x_i$ must be nonzero, and so
  the sets $U_1^\pm, \ldots, U_{n + 1}^\pm$ cover $S^n$. Now for each $U_i^\pm$, we can solve
  for the equation $\abs{ x } = 1$, and find that $x \in S^n \cap U_i^\pm$ iff
  \begin{equation*}
    x_i = \pm \sqrt{1 - \sum_{\substack{j=1 \\ j \neq i}}^{n + 1} x_j^2 }.
  \end{equation*}
  Since the square root is a continuous function, it follows that the intersection of $S^n$ with
  $U_i^\pm$ is the graph of a continuous function. This intersection is therefore locally
  Euclidean, showing to us that $S^n$ is indeed a manifold.
\end{eg}

The following lemma allows us to, essentially, `glue' surfaces to one another.

\begin{lemma}[Glueing Lemma]\index{Glueing Lemma}\label{lemma:glueing_lemma}
  Let $X$ and $Y$ be topological spaces, and let $\{ A_i \}$ be either an arbitrary open cover
  of $X$, or a finite closed cover of $X$. Suppose that we are given continuous maps $f_i : A_i
  \to Y$ that agree on overlaps, i.e.
  \begin{equation*}
    f_i \restriction_{A_i \cap A_j} = f_k \restriction_{A_i \cap A_j}.
  \end{equation*}
  Then there exists a unique continuous map $f : X \to Y$ whose restriction to each $A_i$ is
  equal to $f_i$.
\end{lemma}

With that, we can construct the following space.

\begin{defn}[Adjunction Space]\label{defn:adjunction_space}
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (0, 0) circle(0.7);
      \node at (-0.7, 0.7) {$M$};
      \draw (2, 0) circle(0.7);
      \node at (2.7, 0.7) {$N$};
      \begin{scope}
        \clip (0, 0) circle(0.7);
        \draw (0.5, -0.5) circle(0.5);
        \node at (0,0) {$S_1$};
      \end{scope}
      \begin{scope}
        \clip (2, 0) circle(0.7);
        \draw (1.5, -0.5) circle(0.4);
        \node at (2,0) {$S_2$};
      \end{scope}
      \draw[-latex] (0.4,-0.4) to[bend right] node[midway,below] {$f$} (1.6, -0.4);
    \end{tikzpicture}
    \caption{Glueing subsets of two disjoint spaces.}\label{fig:glueing_subsets_of_two_disjoint_spaces_}
  \end{marginfigure}
  Consider 2 manifolds $M$ and $N$ that are of the same dimension, and let $S_1
  \subseteq M$ and $S_2 \subseteq N$. Let $f : S_1 \to S_2$ be a homeomorphism (cf
  \cref{fig:glueing_subsets_of_two_disjoint_spaces_}). Then we define
  \begin{equation*}
    M \cup_f N := M \coprod N \Big/ \Big\lbrace \substack{a \sim f(a) \\ a \in S_1}
  \end{equation*}
  as the \hlnoteb{adjunction space}, and is said to be formed by \hldefn{attaching}
  \hlnoteb{$M$ to $N$ along $f$}. The map $f$ is called the \hldefn{attaching map}.
\end{defn}

\begin{remark}
  By \cref{lemma:glueing_lemma}, there exists a continuous map between $M$ and $N$, and so this
  allows us to know that this new structure $M \cup_f N$ is indeed a manifold.
\end{remark}

\begin{defn}[Double]\index{Double}\label{defn:double}
  If $M = N$, with the identity map $\id \restriction_{\partial M}$ as a homeomorphism between
  $\partial M$ and $\partial N$, then we call $M \cup_{\id \restriction_{\partial M}} N$ the
  \hlnoteb{double} of $M$.
\end{defn}

\begin{lemma}[Attaching Manifolds along Their Boundaries]\label{lemma:attaching_manifolds_along_their_boundaries}
  Suppose $M$ is an $n$-dimensional manifold with boundary. Then its double
  $M \cup_{\id \restriction_{\partial M}} M$ is a manifold without boundary.
  More generally, if $M_1$ and $M_2$ are manifolds with non-empty boundaries, then there are
  topological embeddings $e : M_1 \cup_h M_2$ and $f : M_1 \cup_h M_2$ whose images are closed
  subsets of $M_1 \cup_h M_2$ satisfying
  \begin{gather*}
    e(M_1) \cup f(M_2) = M_1 \cup_h M_2 \\
    e(M_1) \cap f(M_2) = e(\partial M_1) = f(\partial M_2).
  \end{gather*}
\end{lemma}

The core idea of the proof is illustrated in \cref{fig:attaching_along_the_boundaries}.

\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \draw (1, 1) arc(90:270:0.15 and 1) -- (1, -1) arc(-90:90:0.15 and 1);
    \draw plot [smooth] coordinates {(1, 1) (0.5, 1.1) (0, 1.2) (-1, 0.9) (-1.6, 0.6) (-2, 0) (-1.5, -0.6) (-1, -0.8) (0, -1.1) (1, -1)};
    \draw (-0.7,0.2) to[bend right] (0.3,0.2);
    \draw (-0.6,0.2) to[bend left] (0.2,0.2);
    \begin{scope}
      \clip (1, -1) arc(270:90:0.15 and 1) -- plot [smooth] coordinates {(1, 1) (0.5, 1.1) (0, 1.2) (-1, 0.9) (-1.6, 0.6) (-2, 0) (-1.5, -0.6) (-1, -0.8) (0, -1.1) (1, -1)};
      \draw[fill=blue!60!foreground, fill opacity=0.6] (1, 0) circle(0.4);
    \end{scope}

    \draw (2, 1) arc(90:270:0.15 and 1) -- plot [smooth] coordinates {(2, -1) (2.5, -1.2) (3, -1.25) (3.5, -1.2) (4, -1) (4.5, -0.6) (5, 0) (4.5, 0.6) (3, 1.2) (2, 1)};
    \draw[dashed] (2, 1) arc(90:-90:0.15 and 1);
    \draw (2.6,0) to[bend right] (3.6,0);
    \draw (2.7,0) to[bend left] (3.5,0);
    \begin{scope}
      \clip (2, 1) arc(90:270:0.15 and 1) -- plot [smooth] coordinates {(2, -1) (2.5, -1.2) (3, -1.25) (3.5, -1.2) (4, -1) (4.5, -0.6) (5, 0) (4.5, 0.6) (3, 1.2) (2, 1)};
      \draw[fill=blue!60!foreground, fill opacity=0.6] (1.8, 0) circle(0.4);
    \end{scope}

    \draw[-latex] (0.9, 0) -- (1.8, 0) node[midway,above] {$q$};

    \draw[-latex] (0.7, 0.5) to[bend left] node[midway,above] {$\phi_1$} (6.3,1.5);
    \draw[->] (6, 1.2) -- (8, 1.2);
    \draw[->] (7, 1) -- (7, 2);
    \draw[fill=blue!60!foreground, fill opacity=0.6] (6.5,1.2) arc(180:0:0.5);
    \node at (7.5,2) {$\mathbb{H}^n$};

    \draw[-latex] (2.1, -0.5) to[bend right] node[midway,below] {$\phi_2$} (6.3,-0.8);
    \draw[->] (6, -1.2) -- (8, -1.2);
    \draw[->] (7, -1.4) -- (7, -0.4);
    \draw[fill=blue!60!foreground, fill opacity=0.6] (6.5,-1.2) arc(180:0:0.5);
    \node at (7.5,-0.4) {$\mathbb{H}^n$};

    \draw[-latex] (8.3, 1.5) to[bend left] (10, 1);
    \draw[-latex] (8.3, -0.8) to[bend right] (10, -1);
    \draw[->] (10, 0) -- (12, 0);
    \draw[->] (11, -1) -- (11, 1);
    \draw[fill=blue!60!foreground, fill opacity=0.6] (11, 0) circle(0.5);
  \end{tikzpicture}
  \caption{Attaching along the boundaries}
  \label{fig:attaching_along_the_boundaries}
\end{figure*}

\begin{proof}[Sketch]
  \hlbnotea{Step 1} Find $q$.

  \noindent
  \hlbnotea{Step 2} $q \restriction_(\Int(M_1))$ is an embedding.

  \noindent
  \hlbnotea{Step 3} Define $\phi_1$ and $\phi_2$.

  \noindent
  \hlbnotea{Step 4} Put the two together.

\end{proof}

\newthought{To end} this lecture today, we recalled that we can look at the torus as a
$2$-dimensional rectangle with its sides properly identified. Now if we change the
identification of one of the sides by swapping its orientation, we end up with what is
known as a \hlnotea{M\"{o}bius band}.

\begin{marginfigure}
  \centering
  \includegraphics[width=\marginparwidth]{images/mobius-band1.png}
  \caption{M\"{o}bius band}\label{fig:mobius_band}
\end{marginfigure}

Following that, if we also swap the orientation of the other two sides, we get what is known as
the \hlnotea{Klein Bottle}.

\begin{marginfigure}
  \centering
  \includegraphics[width=\marginparwidth]{images/klein-bottle1.png}
  \caption{Klein Bottle}
  \label{fig:klein_bottle}
\end{marginfigure}

% section topological_embeddings (end)

% chapter lecture_9_jan_25th (end)

\chapter{Lecture 10 Jan 28th}%
\label{chp:lecture_10_jan_28th}
% chapter lecture_10_jan_28th

\section{An Enpasse into Orientability}%
\label{sec:an_enpasse_into_orientability}
% section an_enpasse_into_orientability

We talked about orientability in the last lecture. We shall now introduce a more concrete
way of talking about orientability of a surface.

\begin{defn}[Frame]\index{Frame}\label{defn:frame}
  A \hlnoteb{frame} is a continuous pair of vectors $v_1, v_2$ defined on the entire
  surface such that $v_1$ and $v_2$ are linearly independent.
\end{defn}

\begin{eg}
  Here are two graphical examples of a frame:
  \begin{itemize}
    \item \hlbnotec{Cylinder}
      \begin{figure}[ht]
        \centering
        \begin{tikzpicture}
          \draw (0, 0) -- (2, 0) -- (2, 2) -- (0, 2) -- cycle;
          \node[rotate=90] at (0, 1) {>};
          \node[rotate=90] at (2, 1) {>};
          \draw[->] (0.3, 0.3) -- (0.6, 0.3) node[right] {$v_1$};
          \draw[->] (0.3, 0.3) -- (0.3, 0.6) node[above] {$v_2$};
        \end{tikzpicture}
        \caption{Frame on a Cylinder}
        \label{fig:frame_on_a_cylinder}
      \end{figure}

      Note that if we `move' this frame towards the identified sides, the `direction' of
      which the vectors `point at' remains the same (cf.
      \cref{fig:frame_on_the_identified_sides_of_a_cylinder}).
      \begin{figure}[ht]
        \centering
        \begin{tikzpicture}
          \draw (0, 0) -- (2, 0) -- (2, 2) -- (0, 2) -- cycle;
          \node[rotate=90] at (0, 1) {>};
          \node[rotate=90] at (2, 1) {>};
          \draw (1.85, 0.3) -- (2, 0.3);
          \draw[->] (0, 0.3) -- (0.15,0.3) node[right] {$v_1$};
          \draw[->] (1.85, 0.3) -- (1.85, 0.6) node[above] {$v_2$};
        \end{tikzpicture}
        \caption{Frame on the identified sides of a cylinder}
        \label{fig:frame_on_the_identified_sides_of_a_cylinder}
      \end{figure}

    \item \hlbnotec{M\"{o}bius strip}
      \begin{figure}[ht]
        \centering
        \begin{tikzpicture}
          \draw (0, 0) -- (2, 0) -- (2, 2) -- (0, 2) -- cycle;
          \node[rotate=90] at (0, 1) {>};
          \node[rotate=-90] at (2, 1) {>};
          \draw[->] (0.3, 0.3) -- (0.6, 0.3) node[right] {$v_1$};
          \draw[->] (0.3, 0.3) -- (0.3, 0.6) node[above] {$v_2$};

          \draw (3, 0) -- (5, 0) -- (5, 2) -- (3, 2) -- cycle;
          \node[rotate=90] at (3, 1) {>};
          \node[rotate=-90] at (5, 1) {>};
          \draw (4.9, 0.3) -- (5, 0.3);
          \draw[->,dotted] (5,0.3) -- (5.2,0.3) node[right] {$v_1$};
          \draw[->] (4.9, 0.3) -- (4.9, 0.6) node[above,left] {$v_2$};

          \draw (3,1.7) -- (3.1, 1.7);
          \draw[->,dotted] (3, 1.7) -- (2.8,1.7) node[left] {$v_1$};
          \draw[->] (3.1, 1.7) -- (3.1, 1.4) node[below,right] {$v_2$};
        \end{tikzpicture}
        \caption{Frame on a M\"{o}bius strip}
        \label{fig:frame_on_a_mobius_strip}
      \end{figure}

      The two frames on the second diagram in \cref{fig:frame_on_a_mobius_strip} are the
      same frame. Notice that $v_2$ is now `pointing' in the `opposite direction'
      simultaneously.
  \end{itemize}
\end{eg}

\begin{note}
  The notion of having a `side' for a M\"{o}bius strip makes sense when we embed it in
  $\mathbb{R}^3$.
\end{note}

\begin{eg}\label{eg:projective_space_p2}
  Recall \cref{eg:projective_space}, where we introduced the projective space.
  Since $S^n \simeq \mathbb{R}^{n + 1}$, we can define a projective space with respect to
  $n$-spheres. Let us consider $n = 2$, and consider the group action $\mathcal{G}_2 = \{ -1
  , 1 \}$ given by
  \begin{equation*}
    (-1)(x, y, z) = (-x, -y, -z) \text{ and } (1)(x, y, z) = (x, y, z).
  \end{equation*}
  Then
  \begin{equation*}
    \mathbb{P}^2 = S^2 / \mathcal{G}_2 \simeq \bar{S^+}^2 / \substack{(x, y, z) \sim (-x, -y, -z) \\ (x, y,
    z) \in \partial S^+},
  \end{equation*}
  where $S^+$ is the upper hemisphere.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \pgfmathsetmacro\rad{1};
      \draw (0, 0, 0) circle(\rad);
      \shade[ball color=blue!70!foreground,opacity=0.5] (0, 0, 0) circle(\rad);
      \draw[opacity=0.5] (-\rad,0,0) arc (180:360:{\rad} and {\rad/2});
      \draw[dashed,opacity=0.5] (-\rad,0,0) arc (180:0:{\rad} and {\rad/2});
      \draw[opacity=0.5] (0,\rad,0) arc (90:270:{\rad/2} and \rad);
      \draw[dashed,opacity=0.5] (0,\rad,0) arc (90:-90:{\rad/2} and \rad);

      \fill[color=red,opacity=0.3] (-{\rad}, 0, 0) arc (180:0:{\rad}) -- ({\rad}, 0, 0) arc (360:180:{\rad} and {\rad/2});
      \node[pin={[pin edge=latex'-,pin distance=5pt]45:{$S^+$}}] at (0, 1, 0) {};

      % the axes
      \draw[dashed] (-2, 0, 0) -- (0, 0, 0);
      \draw[dashed] (0, -2, 0) -- (0, 0, 0);
      \draw[dashed] (0, 0, -2) -- (0, 0, 0);
      \draw[-latex'] (0, 0, 0) -- (2, 0, 0) node[right] {$y$};
      \draw[-latex'] (0, 0, 0) -- (0, 2, 0) node[above] {$z$};
      \draw[-latex'] (0, 0, 0) -- (0, 0, 2) node[below,left] {$x$};
    \end{tikzpicture}
    \caption{Projective Plane from $S^2$}\label{fig:projective_plane_from_s_2_}
  \end{marginfigure}
  In a very simple sense, we are `compressing' the sphere by identifying the upper
  hemisphere with the lower hemisphere.

  Note that because of this construction, $\mathbb{P}^2$ is not orientable: a point where
  the vector pointing at a direction parallel to $y$ on the upper hemisphere is identified
  with a point whose corresponding vector points at the opposite direction of $y$.
\end{eg}

\begin{eg}
  Consider the diagram presented in \cref{fig:klein_bottle_as_a_mobius_strip_and_a_disc}.
  \begin{figure*}[ht]
    \centering
    \begin{tikzpicture}
      \draw (0, 0) -- (3, 0) -- (3, 3) -- (0, 3) -- cycle;
      \node[label={90:{$(2)$}}] at (1.5, 3) {>>};
      \node[label={270:{$(2)$}}] at (1.5, 0) {<<};
      \draw[-latex'] (0, 1) -- (3, 1) node[midway,above] {$(1')$};
      \draw[latex'-] (0, 2) -- (3, 2) node[midway,above] {$(1)$};
      \node[rotate=-90,label={90:{$b$}}] at (3, 1.5) {>};
      \node[rotate=90,label={90:{$b$}}] at (0, 1.5) {>};
      \node[left] at (0, 0.5) {$c$};
      \node[left] at (0, 2.5) {$a$};
      \node[right] at (3, 2.5) {$c$};
      \node[right] at (3, 0.5) {$a$};

      \node at (4, 1.5) {$\overset{\text{cut}}{\leadsto}$};

      \draw (5, 1) -- (8, 1) node[midway,below] {$(1)$} -- (8, 1) --
        (8, 3) -- (5, 3) node[midway,above] {$(1')$} -- cycle;
      \draw (5, 2) -- (8, 2) node[midway,label={90:{$(2)$}}] {>>};
      \node[rotate=90,label={90:{$a$}}] at (5, 1.5) {>};
      \node[rotate=-90,label={270:{$a$}}] at (5, 2.5) {>};
      \node[rotate=-90,label={90:{$c$}}] at (8, 1.5) {>};
      \node[rotate=90,label={270:{$c$}}] at (8, 2.5) {>};

      \draw (5, 0) -- (8, 0) node[midway,above] {$(1)$} -- (8, 0) --
        (8, -1) -- (5, -1) node[midway,below] {$(1')$} -- cycle;
      \node[rotate=-90,label={90:{$b$}}] at (8, -0.5) {>};
      \node[rotate=90,label={90:{$b$}}] at (5, -0.5) {>};
      \node at (6.5, -0.5) {M\"{o}bius strip};

      \node at (9, 2.5) {$\leadsto$};

      \draw (11, 2.5) -- (12, 2.5) -- (13, 3) node[midway,rotate=45] {>} --
        (13, 3) arc(0:180:1.5 and 0.75) node[midway,above] {$(1')$} -- (10, 3) -- (11, 2.5)
        node[midway,rotate=-45] {>};
      \draw (12, 2.5) -- (13, 2) node[midway,rotate=-45] {>} --
        (13, 2) arc(360:180:1.5 and 0.75) node[midway,below] {$(1)$} -- (10, 2) -- (11,
        2.5) node[midway,rotate=45] {>};

      \node at (13, 2.5) {$\leadsto$};

      \draw (15.5, 2.5) -- (16, 2.5) node[midway,label={270:{$c$}}] {>} --
        (16, 2.5) arc(0:180:1) node[midway,above] {$(1')$} --
        (14, 2.5) arc(180:360:1) node[midway,below] {$(1)$};
      \draw (14, 2.5) -- (14.5,2.5) node[midway,label={90:{$a$}}] {>};
    \end{tikzpicture}
    \caption{Klein Bottle as a M\"{o}bius Strip and a Disc}
    \label{fig:klein_bottle_as_a_mobius_strip_and_a_disc}
  \end{figure*}
  Now we have a more concrete reason to explain why the Klein Bottle is not embeddable
  into $\mathbb{R}^3$, since the disc itself is not embeddable into $\mathbb{R}^3$.
\end{eg}

\newthought{Recall} \cref{fig:shifted_s_1_for_the_moduli_space} where we saw how $S^1$ is
homeomorphic to the space of all lines in $\mathbb{R}^2$ passing through the origin $(0,
0)$.
\begin{marginfigure}[5pt]
  \centering
  \begin{tikzpicture}
    \draw[->] (-1, 0) -- (3, 0);
    \draw[->] (0, -2) -- (0, 2);
    \draw (1, 0) circle(1);

    \clip (-2, -2) rectangle (3, 2);
    \foreach \m in {-3,-2.5,...,3} {
      \draw[color=blue] plot ({\x}, {\m * \x});
    }
  \end{tikzpicture}
  \caption{Shifted $S^1$ for the moduli space, as shown in
  \cref{fig:shifted_s_1_for_the_moduli_space}.}
\end{marginfigure}

It follows, therefore, that $S^1 \simeq \mathbb{P}^1$.

\begin{quotebox}{green}{foreground}
  What about the space of all lines in $\mathbb{R}^2$?
\end{quotebox}

\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \draw[->] (-2, 0) -- (2, 0);
    \draw[->] (0, -2) -- (0, 2);
    \draw[midarrow={latex}{0.7}] (-2, -2) -- (2, 2) node[right] {$\gamma$};
    \draw[midarrow={latex}{0.7}] (-2, -1) -- (1, 2);
    \draw (0, 0) -- (-0.5, 0.5);
    \draw (-0.4, 0.6) -- (-0.3, 0.5) -- (-0.4, 0.4);
  \end{tikzpicture}
  \caption{Space of all lines in $\mathbb{R}^2$}\label{fig:space_of_all_lines_in_r_2_}
\end{marginfigure}
Observing \cref{fig:space_of_all_lines_in_r_2_}, we see that since we can rotate $\gamma$
at the origin, we may, in particular, rotate $\gamma$ by $180^{\circ}$ to get the opposite
direction. It follows that this space of all lines in $\mathbb{R}^2$ is thus homeomorphic
to $\mathfrak{M}$, a M\"{o}bius strip. Thus, the space of all lines in $\mathbb{R}^2$ is
also not orientable.

\newthought{Following} the same question as above, we may ask:
\begin{quotebox}{green}{foreground}
  What is the space of all lines passing through $(0, 0, 0)$ in $\mathbb{R}^3$?
\end{quotebox}
\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \pgfmathsetmacro\rad{1};
    \draw (0, 0, 0) circle(\rad);
    \shade[ball color=blue!70!foreground,opacity=0.5] (0, 0, 0) circle(\rad);
    \draw[opacity=0.5] (-\rad,0,0) arc (180:360:{\rad} and {\rad/2});
    \draw[dashed,opacity=0.5] (-\rad,0,0) arc (180:0:{\rad} and {\rad/2});
    \draw[opacity=0.5] (0,\rad,0) arc (90:270:{\rad/2} and \rad);
    \draw[dashed,opacity=0.5] (0,\rad,0) arc (90:-90:{\rad/2} and \rad);

    % the axes
    \draw[dashed] (-2, 0, 0) -- (0, 0, 0);
    \draw[dashed] (0, -2, 0) -- (0, 0, 0);
    \draw[dashed] (0, 0, -2) -- (0, 0, 0);
    \draw[-latex'] (0, 0, 0) -- (2, 0, 0) node[right] {$y$};
    \draw[-latex'] (0, 0, 0) -- (0, 2, 0) node[above] {$z$};
    \draw[-latex'] (0, 0, 0) -- (0, 0, 2) node[below,left] {$x$};

    \draw[color=blue] (-2, -2, -2) -- (2, 2, 2);
    \draw[color=blue] (-2, 2, 2) -- (2, -2, -2);
    \draw[color=blue] (1, 2, 0) -- (-1, -2, 0);
    \draw[color=blue] (-2, 0, 1) -- (2, 0, -1);
    \draw[color=blue] (0, 2, -1) -- (0, -2, 1);
    \draw[color=blue] (0, 2, 1) -- (0, -2, -1);
  \end{tikzpicture}
  \caption{Space of all lines passing through $(0, 0, 0)$ in
  $\mathbb{R}^3$}\label{fig:space_of_all_lines_passing_through_r_3}
\end{marginfigure}
It is not difficult to see that this space is homeomorphic to $S^2$, via the map $x
\mapsto \frac{x}{\abs{x}}$ (which is also a quotient map).

Just right before this, we observed, in \cref{eg:projective_space_p2} that
\begin{equation*}
  S^2 / x \sim (-x) \simeq \mathbb{P}^2,
\end{equation*}
where $x \sim (-x)$ via what is called the \hldefn{antipodal map}.

\begin{quotebox}{green}{foreground}
  What about the space of all lines in $\mathbb{R}^3$?
\end{quotebox}

Following a similar observation in \cref{fig:space_of_all_lines_in_r_2_}, we have the
following:

\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
    \rotateRPY{0}{-45}{0};
    \begin{scope}[RPY]
      % the axes
      \draw[dashed] (-2, 0, 0) -- (0, 0, 0);
      \draw[dashed] (0, -2, 0) -- (0, 0, 0);
      \draw[dashed] (0, 0, -2) -- (0, 0, 0);
      \draw[-latex'] (0, 0, 0) -- (4, 0, 0) node[right] {$y$};
      \draw[-latex'] (0, 0, 0) -- (0, 2, 0) node[above] {$z$};
      \draw[-latex'] (0, 0, 0) -- (0, 0, 2) node[below,left] {$x$};
      \shade[opacity=0.5] (-2, 1, -2) -- (4, 1, -2) -- (4, 1, 2) --
        (-2, 1, 2) -- cycle;
      \draw[->] (0, -2, 1) -- (0, 2, -1) node[right] {$\gamma_2$};
      \draw[->] (0, -2, -1) -- (0, 2, 1) node[right] {$\gamma_1$};
      \draw[dashed] (0, 0, 0.5) -- (0, 1, 0.5) -- (0, 1, -0.5) -- (0, 0, -0.5);
      \node[draw,circle,fill,inner sep=1pt,pin={[pin edge={latex'-}, pin
        distance=5pt]30:{$\beta$}}] at (0, 1, -0.5) {};
      \node[draw,circle,fill,inner sep=1pt,pin={[pin edge={latex'-}, pin
        distance=5pt]160:{$\alpha$}}] at (0, 1, 0.5) {};
    \end{scope}
  \end{tikzpicture}
  \caption{Space of all lines in $\mathbb{R}^3$}
  \label{fig:space_of_all_lines_in_r_3}
\end{figure}

Observe that if we rotate $\gamma_1$ and $\gamma_2$ downwards, anchored at $\alpha$ and
$\beta$ respectively, we will eventually get that the two lines are the same line but
pointing at different directions. We see that the space of all lines is also not
orientable.

% section an_enpasse_into_orientability (end)

% chapter lecture_10_jan_28th (end)

\tuftepart{Introduction to Homology}

\chapter{Lecture 11 Jan 30th}%
\label{chp:lecture_11_jan_30th}
% chapter lecture_11_jan_30th

So far, we have constructed and founded several manifolds and came to know about certain
invariances (which we have yet to prove). One may notice that one way to distinguish these
manifolds is by counting the number of `holes' that they have, `holes' of which there is
no continuous deformation that can get rid of it. This brings us into the theory of
\hlnotea{homology}.
\marginnote{I shall be using information both from the class lectures, and from
a video lecture series on Algebraic Topology prepared by
\citealp{njwbalgtopintrohomology}.}

\section{Homology}%
\label{sec:homology}
% section homology

Consider the following three diagrams:

\begin{figure*}[ht]
  \centering
  \begin{tikzpicture}
    \draw (0, 0) circle(1);
    \node[draw,circle,fill,inner sep=1pt] at (0, 1) {};

    \draw (4, 0) arc(0:360:1) -- (4, 0) arc(180:-180:1);
    \node[draw,circle,fill,inner sep=1pt] at (4, 0) {};
    \node[draw,circle,fill,inner sep=1pt] at (3, -1) {};
    \node[draw,circle,fill,inner sep=1pt] at (2.0310876,0.24740396) {};
    \node[draw,circle,fill,inner sep=1pt] at (5, 1) {};
    \node[draw,circle,fill,inner sep=1pt] at (5.9689124,-0.24740396) {};

    \pgfmathsetmacro\fourtyfive{0.7071067811865475244};
    \pgfmathsetmacro\base{8};
    \draw ({\base + \fourtyfive},\fourtyfive) arc(45:315:1) --
      ({\base + \fourtyfive},-\fourtyfive) arc(-135:135:1) --
      ({\base + \fourtyfive},\fourtyfive) --
      ({\base + \fourtyfive},-\fourtyfive);
    \node[draw,circle,fill,inner sep=1pt] at ({\base + \fourtyfive},\fourtyfive) {};
    \node[draw,circle,fill,inner sep=1pt] at ({\base + \fourtyfive},-\fourtyfive) {};
    \node[draw,circle,fill,inner sep=1pt] at ({\base - 1}, 0) {};
    \node[draw,circle,fill,inner sep=1pt] at ({\base + 1 + 2*\fourtyfive}, 0) {};
  \end{tikzpicture}
  \caption{Three different kinds of loops}
  \label{fig:three_different_kinds_of_loops}
\end{figure*}

By inspection, we notice that the leftmost diagram has one `hole', the center
diagram has two, while the rightmost diagram also has two. One way we can
calculate this number of `holes' is by counting the number of loops there are.
We see that in the first and second diagram, the number of loops and the number
of `holes' coincide. However, in the third diagram, there are a total of 3
loops:
\begin{itemize}
  \item $z_1 = \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} = (1 \; 2)(2 \; 3)(3 \;
    1)$;
  \item $z_2 = \begin{pmatrix} 1 & 4 & 3 \end{pmatrix} = (1 \; 4)(4 \; 3)(3 \;
    1)$; and
  \item $z_3 = \begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix} = (1 \; 2)(2 \; 3)(3
    \; 4)(4 \; 1)$.
\end{itemize}
\begin{marginfigure}
  \centering
  \begin{tikzpicture}
    \pgfmathsetmacro\fourtyfive{0.7071067811865475244};
    \pgfmathsetmacro\base{0};
    \draw ({\base + \fourtyfive},\fourtyfive) arc(45:315:1) --
      ({\base + \fourtyfive},-\fourtyfive) arc(-135:135:1) --
      ({\base + \fourtyfive},\fourtyfive) --
      ({\base + \fourtyfive},-\fourtyfive);
    \node[draw,circle,fill,inner sep=1pt,label={90:{$(1)$}}] at ({\base + \fourtyfive},\fourtyfive) {};
    \node[draw,circle,fill,inner sep=1pt,label={180:{$(2)$}}] at ({\base - 1}, 0) {};
    \node[draw,circle,fill,inner sep=1pt,label={270:{$(3)$}}] at ({\base + \fourtyfive},-\fourtyfive) {};
    \node[draw,circle,fill,inner sep=1pt,label={0:{$(4)$}}] at ({\base + 1 + 2*\fourtyfive}, 0) {};
  \end{tikzpicture}
  \caption{Third diagram has three loops}\label{fig:third_diagram_has_three_loops}
\end{marginfigure}
We see that $z_3$ does not entire capture a `hole', since it also captures the
line, or \hlnotea{edge}, $(1) (3)$.

Also, note that $(1 \; 2 \; 3)$ is not a unique representation of $z_1$; we can
also describe $z_1$ as either\sidenote{We can look at the following idea from
group theory when we talk about \hlnotea{transpositions}, i.e. the swapping of
two `indices', for every one of the descriptions. For instance,
\begin{align*}
  (1 \; 2 \; 3) &= (1 \; 2)(2 \; 3)(3 \; 1) \\
                &= -(2 \; 1)(1 \; 3)(3 \; 2) \\
                &= -(2 \; 1 \; 3).
\end{align*}}
\begin{gather*}
  (3 \; 1 \; 2), \, (2 \; 3 \; 1), -(1 \; 3 \; 2), -(3 \; 2 \; 1), \text{ or }
  -(2 \; 1 \; 3).
\end{gather*}

We shall call these expressions as \hldefn{cycles}\sidenote{This is taking a
page out of graph theory.}.

Notice that geometrically the cycle $z_3$ looks like $z_1 - z_2$, which one can
trace the diagram to see that this `minusing' between cycles gives us different
cycles.

To further refine this approach, consider the following diagram:
\begin{figure}[ht]
  \centering
  \begin{tikzpicture}
    \pgfmathsetmacro\fourtyfive{0.7071067811865475244};
    \draw (0, 0) circle(1);
    \node[draw,circle,fill,inner sep=1pt,label={135:{$(a)$}}] at (0, 1) {};
    \node[draw,circle,fill,inner sep=1pt] at ({0 - \fourtyfive}, -\fourtyfive)
      {};
    \node[draw,circle,fill,inner sep=1pt] at (\fourtyfive, -\fourtyfive) {};
    \draw (0, 1) to[bend left] node[midway,above] {$e$} (1, 2)
      node[draw,circle,fill,inner sep=1pt,label={0:{$(b)$}}] {};
  \end{tikzpicture}
  \caption[][30pt]{Simple diagram with a branching edge}
  \label{fig:simple_diagram_with_a_branching_edge}
\end{figure}

Notice that $(a \; b)(b \; a)$ also goes back to the same point, forming, in a
sense, a closed path. However, this does not enclose a `hole', despite being
algebraically similar to the cycles that we have described above. It is
important that we find a way to make this distinction.

\newthought{To make this} distinction, we look at the \hldefn{boundary} of the
edges, which is defined with respect to the \hlnotea{vertices} of an edge.

\begin{defn}[Boundary of an Edge]\index{Boundary}\label{defn:boundary_of_an_edge}
  We define the boundary of an edge $e$, whose endpoints, or more correctly
  vertices, are $a$ and $b$, as follows:
  \begin{itemize}
    \item if $e$ is an undirected edge, then both $a$ and $b$ are the boundaries
      of $e$;
    \item if $e$ is a directed edge, say from $a$ to $b$, then we denote the
      boundary of $e$ as
      \begin{equation*}
        \partial e = b - a.
      \end{equation*}
  \end{itemize}
\end{defn}

\begin{remark}
  There is a good reason behind using notation $\partial e = b - a$, which we
  shall see later.
\end{remark}

\begin{eg}
  Referring back to \cref{fig:simple_diagram_with_a_branching_edge}, we see that
  the boundary of $e$ is
  \begin{equation*}
    \partial e = (b) - (a).
  \end{equation*}
\end{eg}

\begin{eg}\label{eg:identifying_actual_cycles}
  Looking back at \cref{fig:third_diagram_has_three_loops}, we have that
  \begin{align*}
    \partial (1 \; 2) &= (2) - (1) & \partial (2 \; 3) &= (3) - (2) \\
    \partial (3 \; 1) &= (1) - (3) & \partial (1 \; 4) &= (4) - (1) \\
    \partial (4 \; 3) &= (3) - (4).
  \end{align*}

  Notice that
  \begin{equation*}
    \partial(1 \; 2) + \partial(2 \; 3) + \partial(3 \; 1)
    = (2) - (1) + (3) - (2) + (1) - (3) = 0,
  \end{equation*}
  and $(1 \; 2)(2 \; 3)(3 \; 1)$ is one of the cycles which we have identified
  that encloses a `hole'.
\end{eg}

For reasons that will become clear later, we shall, from hereon, write
\begin{equation*}
  (1 \; 2)(2 \; 3)(3 \; 1) = (1 \; 2) + (2 \; 3) + (3 \; 1)
\end{equation*}
to represent the cycle $z_1$ in \cref{fig:third_diagram_has_three_loops}.

\newthought{Let} $C_0$ denote the \hlnotea{free abelian group}\sidenote{See
\cref{sec:free_abelian_groups}} on vertices, and $C_1$ denote the free abelian
group on (directed) edges.

\begin{eg}
  Let $C_0$ be the free abelian group on the vertices $x, y, z$. Then an element
  of $C_0$ would be $2x + 3y - z$, i.e. $C_0$ is a group that contains linear
  combinations of $x, y, z$ with integer coefficients.

  Following the same `rule', elements of $C_1$, the free abelian group on the
  directed edges $e, d, f$ are of the form $\alpha_1 e + \alpha_2 d + \alpha_3
  f$.
\end{eg}

We shall now give $C_n$ a proper name.

\begin{defn}[$n$-Chain]\index{$n$-Chain}\label{defn:n_chain}
  Elements of $C_0$ are called \hlnoteb{$0$-dimensional chains}, while elements
  of $C_1$ are called \hlnoteb{$1$-dimensional chains}.

  In general, elements of $C_n$ are called \hlnoteb{$n$-dimensional chains}, or
  simply $n$-chains. $C_n$ itself is called a \hlnoteb{group of $n$-chains}.
\end{defn}

We may now make the definition of $\partial$ more formally so:

\begin{defn}[Boundary of Chains]\index{Boundary of Chains}\label{defn:boundary_of_chains}
  The \hlnoteb{boundary of an $n$-chain}, denoted $\partial e$, where $e$ is an
  $n$-chain is a \hlnotea{homomorphism}
  \begin{equation*}
    \partial : C_n \to C_{n - 1}.
  \end{equation*}
\end{defn}

\begin{remark}
  Since $\partial$ is a homomorphism, it is therefore linear. One can verify
  that the group of $n$-chains is indeed an abelian (commutative) group, which
  explains the notation $e + d$.

  Putting the two together, we have
  \begin{equation*}
    \partial(e + d) = \partial e + \partial d.
  \end{equation*}
\end{remark}

\newthought{Following} the observation that we noticed in
\cref{eg:identifying_actual_cycles}, one can go back on our examples and verify
to oneself that the edges that form a cycle will have no boundary, i.e. if $e +
d + f$ forms a cycle, then $\partial(e + d + f) = 0$.

We shall now try to find all the possible cycles in the following simple
example:

\begin{eg}\label{eg:finding_the_number_of_cycles}
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \pgfmathsetmacro\fourtyfive{0.7071067811865475244};
      \draw (0, 0) circle(1);
      \node[draw,circle,fill,inner sep=1pt,label={135:{$(1)$}}] at (0, 1) {};
      \node[draw,circle,fill,inner sep=1pt,label={135:{$(2)$}}] at ({0 -
        \fourtyfive}, -\fourtyfive) {};
      \node[draw,circle,fill,inner sep=1pt,label={135:{$(3)$}}] at (\fourtyfive,
        -\fourtyfive) {};
    \end{tikzpicture}
    \caption{Simple example into using boundaries to identify cycles}\label{fig:simple_example_into_using_boundaries_to_identify_cycles}
  \end{marginfigure}
  Consider \cref{fig:simple_example_into_using_boundaries_to_identify_cycles}.
  We want to solve for
  \begin{equation*}
    \partial [ \alpha_1 (1 \; 2) + \alpha_2 (2 \; 3) + \alpha_3 (3 \; 1) ] = 0.
  \end{equation*}
  Then we have
  \begin{gather*}
    \alpha_1 [(2) - (1)] + \alpha_2 [(3) - (2)] + \alpha_3 [(1) - (3)] = 0 \\
    (1) (\alpha_3 - \alpha_1) + (2) (\alpha_1 - \alpha_2) + (3) (\alpha_2 -
    \alpha_3) = 0.
  \end{gather*}
  We thus have the following system of equations
  \begin{gather*}
    -\alpha_1 + \alpha_3 = 0 \\
    \alpha_1 - \alpha_2 = 0 \\
    \alpha_2 - \alpha_3 = 0.
  \end{gather*}
  We shall solve for this using \hlnotea{row reduction} on matrices: we have
  \begin{align*}
    \begin{bmatrix}
      -1 & 0 & 1 \\
      1 & -1 & 0 \\
      0 & 1 & -1
    \end{bmatrix} \sim
    \begin{bmatrix}
      1 & 0 & -1 \\
      0 & 1 & -1 \\
      0 & 0 & 0
    \end{bmatrix}.
  \end{align*}
  By letting $\alpha_3 = r$, we have that the solutions are
  \begin{equation*}
    \begin{pmatrix}
      \alpha_1 \\ \alpha_2 \\ \alpha_3
    \end{pmatrix} =
    r \begin{pmatrix}
      1 \\ 1 \\ 1
    \end{pmatrix}.
  \end{equation*}
  It follows that the null space for this solution is spanned by the vector
  $\left( \begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix} \right)$, which is
  therefore isomorphic to $\mathbb{Z}$.
  \marginnote{\cref{eg:finding_the_number_of_cycles} gives us a procedure for
  finding the number of cycles:
  \begin{enumerate}
    \item Equate the boundary of an arbitrary chain to $0$.
    \item Find the solutions to the system of equations, which is just finding
      the \hlnotea{null space} to the matrix that the system of equations
      coincide with.
  \end{enumerate}}
\end{eg}

\begin{remark}
  It is no coincidence that the dimension of $\mathbb{Z}$ agrees with the total
  number of cycles.
\end{remark}

We shall now make the following definition:

\begin{defn}[Cycle]\index{Cycle}\label{defn:cycle}
  A chain $e$ is a \hlnoteb{cycle} if $\partial e = 0$.
\end{defn}

% section homology (end)

% chapter lecture_11_jan_30th (end)

\chapter{Lecture 12 Feb 01st}%
\label{chp:lecture_12_feb_01st}
% chapter lecture_12_feb_01st

\section{Homology (Continued)}%
\label{sec:homology_continued}
% section homology_continued

Continuing from before, we may classify chains into equivalence classes if the
chains form a boundary.

\begin{defn}[Homologous]\index{Homologous}\label{defn:homologous}
  Two $n$-chains $a_1$, $a_2$ are said to be \hlnoteb{homologous} if
  \begin{equation*}
    a_1 - a_2 \in \Img \partial_{n + 1},
  \end{equation*}
  where $\partial_{n + 1}$ is the boundary operator from $C_{n + 1}$ to $C_n$.
  We shall write $a_1 \sim a_2$.

  Two $n$-cycles $z_1$, $z_2$ are said to be \hlnoteb{homologous} if
  \begin{equation*}
    z_1 - z_2 = \partial_{n + 1} \sigma
  \end{equation*}
  for some $(n + 1)$-chain, and we also label this as $z_1 \sim z_2$.
\end{defn}

\begin{eg}\label{eg:first_eg_of_triangulation}
  Consider a disc in $\mathbb{R}^2$. We know that the disc is homeomorphic to a
  triangle whose interior is filled. This process of identifying a `shape' with
  triangles is called \hldefn{triangulation}.
  \begin{figure*}[ht]
    \centering
    \begin{tikzpicture}
      \draw[fill=blue,fill opacity=0.5] (0, 0) circle(1);
      \node at (2, 0) {$\simeq$};
      \draw[fill=blue,fill opacity=0.5] (4, 1) -- (3, -1) node[midway,left]
        {$b_1$} -- (5, -1) node[midway,below] {$b_2$} -- cycle
        node[midway,right] {$b_3$};
      \node[draw,circle,fill,inner sep=1pt,label={90:{$v_1$}}] at (4, 1) {};
      \node[draw,circle,fill,inner sep=1pt,label={225:{$v_2$}}] at (3, -1) {};
      \node[draw,circle,fill,inner sep=1pt,label={315:{$v_3$}}] at (5, -1) {};
      \node[pin={[pin edge={latex'-}, pin distance=15pt]30:{$2$-dimensional
        face}}] at (4, 0) {$a$};
    \end{tikzpicture}
    \caption{Triangulation of a disc}
    \label{fig:triangulation_of_a_disc}
  \end{figure*}
  We see that
  \begin{equation*}
    b_1 \sim (b_2 + b_3) \quad b_2 \sim (b_3 + b_1) \quad b_3 \sim (b_1 + b_2),
  \end{equation*}
  which forms the boundaries of $a$, and
  \begin{equation*}
    v_1 \sim v_2 \quad v_2 \sim v_3 \quad v_3 \sim v_1,
  \end{equation*}
  which forms the boundary of $b_1$, $b_2$ and $b_3$ respectively.  However, $a$
  is not homeomorphic to any other \hlnotea{faces}.
\end{eg}

\begin{eg}
  Similar to the above, we can triangulate a sphere using a crystal-like shape
  (cf. \cref{fig:triangulation_of_a_sphere}).
  \begin{figure*}[ht]
    \centering
    \begin{tikzpicture}
      \pgfmathsetmacro\rad{1};
      \draw (0, 0, 0) circle(\rad);
      \draw[opacity=0.5] (-\rad,0,0) arc (180:360:{\rad} and {\rad/2});
      \draw[dashed,opacity=0.5] (-\rad,0,0) arc (180:0:{\rad} and {\rad/2});
      \draw[opacity=0.5] (0,\rad,0) arc (90:270:{\rad/2} and \rad);
      \draw[dashed,opacity=0.5] (0,\rad,0) arc (90:-90:{\rad/2} and \rad);

      \node at (2, 0) {$\simeq$};

      \draw (4, 0, 1) -- (3, 0, 0) -- (4, 1, 0) -- (5, 0, 0) -- (4, 0, 1) -- (4,
        1, 0);
      \draw (3, 0, 0) -- (4, -1, 0) -- (5, 0, 0);
      \draw (4, -1, 0) -- (4, 0, 1);
      \draw[dashed] (3, 0, 0) -- (5, 0, 0);
    \end{tikzpicture}
    \caption{Triangulation of a sphere}
    \label{fig:triangulation_of_a_sphere}
  \end{figure*}
\end{eg}

Under the maps $\partial_n$, for $n \geq 0$, we can relate the chains as
follows:

\begin{defn}[Chain Complex]\index{Chain Complex}\label{defn:chain_complex}
  A sequence of groups of chains
  \begin{equation*}
    \hdots \overset{\partial_{n + 1}}{\to} C_n \overset{\partial_n}{\to} C_{n -
    1} \overset{\partial_{n - 1}}{\to} \hdots \overset{\partial_0}{\to} 0
  \end{equation*}
  is called a \hlnoteb{chain complex} if $\partial_{i} \circ \partial_{i + 1} =
  0$, for $i \geq 0$.
\end{defn}

\marginnote{Note that the $n$-chains are constructed using a specific
triangulation. However, it does not matter, which is something that shall be
proven much later.}
\begin{eg}
  In \cref{eg:first_eg_of_triangulation}, we have the following chain complex
  \begin{equation*}
    \hdots \to C_3 = 0 \overset{\partial_3}{\to} C_2 \overset{\partial_2}{\to}
    C_1 \overset{\partial_1}{\to} C_0 \overset{\partial_0}{\to} 0,
  \end{equation*}
  where we note that $C_n = 0$ for $n \geq 3$ since there are no $3$-chains and
  above, we have
  \begin{align*}
    C_2 &= \left\{ na \mmid n \in \mathbb{Z} \right\} \simeq \mathbb{Z} \\
    C_1 &= \left\{ \sum_{i=1}^{3} n_i b_i \mid n_i \in \mathbb{Z} \right\}
      \simeq \mathbb{Z}^3 \\
    C_0 &= \left\{ \sum_{i=1}^{3} n_i a_i \mid n_i \in \mathbb{Z} \right\}
      \simeq \mathbb{Z}^3
  \end{align*}
\end{eg}

We are now ready to provide a more specific definition of \cref{defn:cycle} that
is more useful to us.

\begin{defn}[$k$-cycles]\index{$k$-cycles}\label{defn:_k_cycles}
  A \hlnoteb{$k$-cycle} is a $k$-chain $z$ such that $\partial_k z = 0$, i.e.
  $z \in \ker \partial_k$. We denote the set of $k$-cycles as $Z_k$.
\end{defn}

Note that chains are \hlnotea{free abelian groups}, and so $Z_k \triangleleft
C_k$. By the property $\partial_k \circ \partial_{k + 1}$, we have that $\Img
\partial_{k + 1} \triangleleft Z_k$.

\begin{defn}[Homology Groups]\index{Homology Groups}\label{defn:homology_groups}
  The \hldefn{$k$-th Homology} is defined as
  \begin{equation*}
    H_k := \faktor{\ker \partial_k}{\Img \partial_{k + 1}}.
  \end{equation*}
\end{defn}

\newthought{Now why} is it true that $\partial_k \circ \partial_{k + 1} = 0$?

\begin{propo}[Boundary of a Boundary is Zero]\label{propo:boundary_of_a_boundary_is_zero}
  For any $k \geq 0$, we have
  \begin{equation*}
    \partial_k \circ \partial_{k + 1} = 0.
  \end{equation*}
\end{propo}

We shall prove this for the $k = 1$ case. One can prove for the more general
case later on using \hlnotea{simplicial homology}.

\begin{proof}
  Consider the $2$-face $[(1) \; (2) \; (3)]$ as shown in
  \cref{fig:your_everyday_2_face}.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (0, 1) -- (-1, 0) -- (1, 0) -- cycle;
      \node[above] at (0, 1) {$(1)$};
      \node[below,left] at (-1, 0) {$(2)$};
      \node[below,right] at (1, 0) {$(3)$};
      \node at (0, 0.4) {$\circlearrowleft$};
    \end{tikzpicture}
    \caption{Your everyday $2$-face}\label{fig:your_everyday_2_face}
  \end{marginfigure}
  We have
  \begin{align*}
    \partial_2 [(1) \; (2) \; (3)] = [ (3) \; (1) ] + [ (1) \; (2) ] + [ (2) \;
    (3)].
  \end{align*}
  Then
  \begin{align*}
    &\partial_1 ([(3) \; (1)] + [(1) \; (2)] + [(2) \; (3)]) \\
    &= (1) - (3) + (2) - (1) + (3) - (2) = 0.
  \end{align*}
\end{proof}

% section homology_continued (end)

% chapter lecture_12_feb_01st (end)

\chapter{Lecture 13 Feb 04th}%
\label{chp:lecture_13_feb_04th}
% chapter lecture_13_feb_04th

\marginnote{I missed this lecture and I have not been able to obtain any notes
for it. This lecture seems to have introduced simplicial complexes, simplices,
and did some homology on some of the basic structures that we have came across.
I will try to fill these in.}

% chapter lecture_13_feb_04th (end)

\chapter{Lecture 14 Feb 08th}%
\label{chp:lecture_14_feb_08th}
% chapter lecture_14_feb_08th

\section{Simplicial Homology (Continued)}%
\label{sec:simplicial_homology_continued}
% section simplicial_homology_continued

In the last lecture, we showed the the following:

\begin{itemize}
  \item The homology group of the filled triangle \tikz\draw[fill=blue] (0, 0.2) --
    (-0.2, -0.2) -- (0.2, -0.2) -- (0, 0.2); is
    \begin{align*}
      H_0(\tikz\draw[fill=blue] (0, 0.2) -- (-0.2,-0.2) -- (0.2, -0.2) -- (0,
      0.2);) &= \mathbb{Z} \\
      H_1(\tikz\draw[fill=blue] (0, 0.2) -- (-0.2,-0.2) -- (0.2, -0.2) -- (0,
      0.2);) &= 0 \\
      H_2(\tikz\draw[fill=blue] (0, 0.2) -- (-0.2,-0.2) -- (0.2, -0.2) -- (0,
      0.2);) &= 0
    \end{align*}

  \item The homology group of the \hldefn{tetrahedron} \tikz\draw (0, 0.2, 0) --
    (-0.2, 0, 0.2) -- (0.1, 0, 0.2) -- (0, 0.2, 0) -- (0, 0, -0.2) -- (0.1, 0,
    0.2) -- (-0.2, 0, 0.2) -- (0, 0, -0.2);\ is
    \begin{align*}
      H_0(\tikz\draw (0, 0.2, 0) -- (-0.2, 0, 0.2) -- (0.1, 0, 0.2) -- (0, 0.2,
      0) -- (0, 0, -0.2) -- (0.1, 0, 0.2) -- (-0.2, 0, 0.2) -- (0, 0, -0.2);) &=
      \mathbb{Z} \\
      H_1(\tikz\draw (0, 0.2, 0) -- (-0.2, 0, 0.2) -- (0.1, 0, 0.2) -- (0, 0.2,
      0) -- (0, 0, -0.2) -- (0.1, 0, 0.2) -- (-0.2, 0, 0.2) -- (0, 0, -0.2);) &=
      0 \\
      H_3(\tikz\draw (0, 0.2, 0) -- (-0.2, 0, 0.2) -- (0.1, 0, 0.2) -- (0, 0.2,
      0) -- (0, 0, -0.2) -- (0.1, 0, 0.2) -- (-0.2, 0, 0.2) -- (0, 0, -0.2);) &=
      \mathbb{Z}
    \end{align*}
\end{itemize}

We shall now turn to studying the homology groups of the torus.

\begin{eg}[Homology of the Torus]\label{eg:homology_of_the_torus}
  Our goal is to show that the homology group of the torus $\mathbb{T}^2$, as shown in
  \cref{fig:homology_of_a_torus}, is given by
  \begin{align*}
    H_0(\mathbb{T}^2) &= \mathbb{Z} \\
    H_1(\mathbb{T}^2) &= \Span_{\mathbb{Z}} \{ \tikz\draw (0, -0.5) arc(-90:270:0.15 and
    0.2);, \tikz\draw (0.35, 0) arc(0:360:0.35 and 0.2); \}
           = \mathbb{Z} \oplus \mathbb{Z} \\
    H_2(\mathbb{T}^2) &= \mathbb{Z}
  \end{align*}
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (1, 0) arc(0:360:1 and 0.5);
      \draw (0.3, 0) arc(360:180:0.3 and 0.1);
      \draw (0.25, 0) arc(0:180:0.25 and 0.05);
      \draw[dashed] (1, 0) arc(0:180:1 and 0.25);
      \draw (1, 0) arc(360:180:1 and 0.35);
      \draw (0, -0.5) arc(270:90:0.15 and 0.2);
      \draw[dashed] (0, -0.5) arc(-90:90:0.15 and 0.2);
    \end{tikzpicture}
    \caption{Homology of a torus}\label{fig:homology_of_a_torus}
  \end{marginfigure}
  We may triangulate $\mathbb{T}^2$ as in \cref{fig:triangulating_t_2_}.
  \begin{figure*}[ht]
    \centering
    \begin{tikzpicture}
      \draw (0, 3) -- (0, 2) node[left] {$a_1$} -- (0, 1) node[left] {$a_2$}
        -- (0, 0) node[left] {$a_0$} -- (1, 0) node[below] {$b_0$} --
        (2, 0) node[below] {$c_0$} -- (3, 0) node[right] {$a_0$} --
        (3, 1) node[right] {$a_2$} -- (3, 2) node[right] {$a_1$} --
        (3, 3) node[right] {$a_0$} -- (2, 3) node[above] {$c_0$} --
        (1, 3) node[above] {$b_0$} -- (0, 3) node[left] {$a_0$};
      \draw (0, 2) -- (1, 2) node[left=7pt,above] {$b_1$} --
        (2, 2) node[left=7pt,above] {$c_1$} -- (3, 2);
      \draw (0, 1) -- (1, 1) node[left=7pt,above] {$b_2$} --
        (2, 1) node[left=7pt,above] {$c_2$} -- (3, 1);
      \draw (1, 0) -- (1, 3);
      \draw (2, 0) -- (2, 3);
      \foreach \x in {0,1,2} {
      \foreach \y in {0,1,2} {
        \draw (\x, \y) -- ($(\x, \y) + (1,1)$);
      }
      }
      \node[rotate=90,color=red] at (0, 1.5) {>};
      \node[rotate=90,color=red] at (3, 1.5) {>};
      \node[color=red] at (1.5, 0) {>>};
      \node[color=red] at (1.5, 3) {>>};

      \node at (0.8,0.2) {$\circlearrowleft$};
      \node at (0.9, 0.6) {$t_1$};
      \node[rotate=45,color=blue] at (0.5, 0.5) {>};
      \node at (0.4, 0.7) {$e_1$};
    \end{tikzpicture}
    \caption{Triangulating $\mathbb{T}^2$}
    \label{fig:triangulating_t_2_}
  \end{figure*}

  Note that the red arrows and blue arrows indicate different things:
  \begin{itemize}
    \item \textcolor{red}{>} indicates the sides which are identified with one
      another;
    \item \textcolor{blue}{>} indicates an orientation on the $1$-chain.
  \end{itemize}

  \hlbnoted{$H_0(\mathbb{T}^2)$} Notice that there are a total of $10$ vertices
  in \cref{fig:triangulating_t_2_}. Thus $\ker \partial_0 = \mathbb{Z}^{10}$.

  Notice that one may fill in the orientations on the $1$-chains in a way
  illustrated in \cref{fig:an_set_of_orientations_for_the_1_chains_on_t_2}.
  \begin{figure}[ht]
    \centering
    \begin{tikzpicture}
      \draw (0, 3) -- (0, 2) node[left] {$a_1$} -- (0, 1) node[left] {$a_2$}
        -- (0, 0) node[left] {$a_0$} -- (1, 0) node[below] {$b_0$} --
        (2, 0) node[below] {$c_0$} -- (3, 0) node[right] {$a_0$} --
        (3, 1) node[right] {$a_2$} -- (3, 2) node[right] {$a_1$} --
        (3, 3) node[right] {$a_0$} -- (2, 3) node[above] {$c_0$} --
        (1, 3) node[above] {$b_0$} -- (0, 3) node[left] {$a_0$};
      \draw (0, 2) -- (1, 2) node[left=7pt,above] {$b_1$} --
        (2, 2) node[left=7pt,above] {$c_1$} -- (3, 2);
      \draw (0, 1) -- (1, 1) node[left=7pt,above] {$b_2$} --
        (2, 1) node[left=7pt,above] {$c_2$} -- (3, 1);
      \draw (1, 0) -- (1, 3);
      \draw (2, 0) -- (2, 3);
      \foreach \x in {0,1,2,3} {
      \foreach \y in {0,1,2} {
        \ifthenelse{\x < 3}{
          \draw (\x, \y) -- ($(\x, \y) + (1,1)$) node[midway,rotate=45,color=blue] {>};
        }{}
        \node[rotate=-90,color=blue] at ($(\x, \y) + (0, 0.5)$) {>};
        \node[color=blue] at ($(\y, \x) + (0.5, 0)$) {<};
      }
      }
      \node[rotate=45,color=blue] at (0.5, 0.5) {>};
    \end{tikzpicture}
    \caption{An set of orientations for the $1$-chains on $\mathbb{T}^2$}
    \label{fig:an_set_of_orientations_for_the_1_chains_on_t_2}
  \end{figure}
  It can be shown that given any linear combination of $1$-chains, one of the
  coefficients will end up being dependent on all others. In other words, $\Img
  \partial_1 = \mathbb{Z}^9$. Thus $H_0(\mathbb{T}^2) = \mathbb{Z}$, as
  required.

  \hlbnoted{$H_2(\mathbb{T}^2)$} Note that $C_3 = 0$, and so $\Img \partial_3 =
  0$. By the triangulation in \cref{fig:triangulating_t_2_}, we know that $C_2
  \simeq \mathbb{Z}^{18}$. Now by applying the same orientation on each of the
  $2$-faces, we notice that in order for the $1$-cycles to cancel out each other
  (i.e. in order for $\partial_2 \Delta$, where $\Delta$ is an arbitrary
  $2$-chain, to be $0$), we need the coefficients of all of the $2$-chains to be
  the same. Thus $\ker \partial_2 = \mathbb{Z}$. It follows that
  $H_2(\mathbb{T}^2) = \mathbb{Z}$.

  \hlbnoted{$H_1(\mathbb{T}^2)$} \sidenote[][-100pt]{This is extremely tedious to deal
  with if we go down the normal route of calculating the nullity and rank. We
  shall use the following procedure to arrive at $H_1(\mathbb{T}^2)$.
  \begin{procedure}[Figuring out $H_1(\mathbb{T}^2)$]
    \begin{enumerate}
      \item Show that every $1$-cycle is \hyperref[defn:homologous]{homologous}
        to the $1$-cycle that lies entirely on the boundary of the rectangle in
        \cref{fig:triangulating_t_2_}.
      \item Show that every $1$-cycle supported on the boundary of the rectangle
        is of the form $m_1 z_1 + m_2 z_2$, where $m_1, m_2 \in \mathbb{Z}$ and
        $z_1, z_2$ are as labelled in
        \cref{fig:_z_1_and_z_2_on_t_2_as_a_basis_for_the_null_space_of_1_chains_}.
      \item Show that $z_1$ and $z_2$ are not homologous.
    \end{enumerate}
  \end{procedure}
  }
  \begin{figure*}[ht]
    \centering
    \begin{tikzpicture}
      \draw (3, 0) arc(0:360:3 and 1.5);
      \draw (1, 0) arc(360:180:1 and 0.25);
      \draw (0.8, -0.15) arc(0:180:0.8 and 0.15);
      \draw[color=magenta] (3, 0) arc(360:180:3 and 1) node[near end,above]
        {$z_2$};
      \draw[dashed,color=magenta] (3, 0) arc(0:180:3 and 1);
      \draw[color=cyan] (0, -1.5) arc(270:90:0.3125 and 0.625) node[near end,
        left] {$z_1$};
      \draw[dashed,color=cyan] (0, -1.5) arc(-90:90:0.3125 and 0.625);
    \end{tikzpicture}
    \caption{$z_1$ and $z_2$ on $\mathbb{T}^2$, as a basis for the null space of
    $1$-chains.}
    \label{fig:_z_1_and_z_2_on_t_2_as_a_basis_for_the_null_space_of_1_chains_}
  \end{figure*}

  \hlwarn{I will fill in the details later when I've worked it out. Some steps
  are probably wrong or not well-justified.}
\end{eg}

\begin{eg}[Homology of the Klein Bottle]\label{eg:homology_of_the_klein_bottle}
  \hlwarn{I will have to revisit the compution of homology.}
\end{eg}

% section simplicial_homology_continued (end)

% chapter lecture_14_feb_08th (end)

\chapter{Lecture 15 Feb 11th}%
\label{chp:lecture_15_feb_11th}
% chapter lecture_15_feb_11th

\section{Free Abelian Groups}%
\label{sec:free_abelian_groups}
% section free_abelian_groups

We shall make a little detour and look at free abelian groups, and look at some
of the results that are useful to us.

\begin{defn}[Free Abelian Groups]\index{Free Abelian Groups}\label{defn:free_abelian_groups}
  A group $G$ is called a \hlnoteb{free abelian group} if every $g \in G$ can be
  uniquely written as a finite sum
  \begin{equation*}
    g = \sum n_\alpha g_\alpha,
  \end{equation*}
  where $n_\alpha$ are (integer) coefficients, and $\{ g_\alpha \}$ is a
  \hlnotea{basis} for the group.
\end{defn}

\begin{eg}
  Consider
  \begin{equation*}
    \tilde{G} = \bigoplus_{j \in \mathbb{N}} \mathbb{Z},
  \end{equation*}
  a sequence of finitely many non-zero components. As a set, $\tilde{G}$ is the
  \hlnotea{infinite Cartesian product} of infinitely many copies of
  $\mathbb{Z}$. We may thus consider the basis elements of $\tilde{G}$ as
  \begin{equation*}
    g_j = (0, \ldots, 0, \overarrow{1}{j-\text{th component}}, 0, \ldots).
  \end{equation*}
\end{eg}

\begin{defn}[Generators of a Group]\index{Generators of a Group}\label{defn:generators_of_a_group}
  If we remove the condition of unique representation, i.e. for $g \in G$,
  \begin{equation*}
    g = \sum n_\alpha g_\alpha
  \end{equation*}
  not necessarily uniquely, we then call $\{ g_\alpha \}$ a set of
  \hlnoteb{generators} for $G$.
\end{defn}

\begin{defn}[Finitely Generated]\index{Finitely Generated}\label{defn:finitely_generated}
  We say that a group $G$ is \hlnoteb{finitely generated} if it has a finite set
  of generators.
\end{defn}

\begin{eg}
  Consider $\tilde{G}$ as in the last example. We have that $\tilde{G}$ is not
  finitely generated.
\end{eg}

\begin{eg}
  $\mathbb{Z}^2$ is finitely generated: its generators are
  \begin{equation*}
    \left\{ g_1 = (1, 0),\,g_2 = (0, 1) \right\}.
  \end{equation*}
\end{eg}

\begin{note}
  Note that the above two examples are free abelian groups. This means that free
  abelian groups \hlimpo{do not have to be finitely generated}.
\end{note}

\begin{ex}
  Show that any subgroup of a finitely generated abelian group is finitely
  generated.
\end{ex}

\begin{proof}
  Let $G$ be a finitely generated abelian group, with generators $\{ g_i
  \}_{i=1}^{n}$. Let  $H \leq G$. Then $\forall h \in H$, since $h \in G$, we
  have that
  \begin{equation*}
    h = \sum n_i g_i
  \end{equation*}
  for some coefficients $n_i$. Thus $H$ is finitely generated.
\end{proof}

\begin{note}
  Observe that if $\{g_\alpha\}$ and $\{h_\alpha\}$ are two finite basis for the
  finitely generated group $G$, then
  \begin{equation*}
    \abs{\{g_\alpha\}} = \abs{\{h_\alpha\}}.
  \end{equation*}
\end{note}

\begin{defn}[Order]\index{Order}\label{defn:order}
  An element $g \in G$ is said to have \hlnoteb{order} $p$ if
  \begin{equation*}
    p \cdot g = 0
  \end{equation*}
  and $\forall q$, if $q \cdot g = 0$, then $p \leq q$.
\end{defn}

\begin{defn}[Torsion Group]\index{Torsion Group}\label{defn:torsion_group}
  A group $G$ is said to be \hlnoteb{torsion} if all of its elements have finite
  order.
\end{defn}

\begin{eg}
  Finite groups are necessarily torsion.
\end{eg}

\begin{ex}
  Let $G$ be an abelian group and
  \begin{equation*}
    H \coloneqq \left\{ g \in G : g \text{ has finite order } \right\}.
  \end{equation*}
  Show that $H < G$. We say that $H$ is a \hldefn{torsion subgroup} of $G$.
\end{ex}

\begin{thm}[Fundamental Theorem of Finitely Generated Abelian Groups]\index{Fundamental Theorem of Finitely Generated Abelian Groups}\label{thm:fundamental_theorem_of_finitely_generated_abelian_groups}
  Let $G$ be a finitely generated abelian group, and $T$ a torsion subgroup.
  Then $G$ \hlnotea{splits}, i.e.
  \begin{equation*}
    G \simeq K \oplus T,
  \end{equation*}
  where $K$ is a \hyperref[defn:free_abelian_groups]{free abelian group}.
  Furthermore,
  \begin{equation*}
    T \simeq T_1 \oplus \hdots \oplus T_k,
  \end{equation*}
  where $T_j$ is of order $m_j$ and $m_1 \mid m_2 \mid \hdots \mid m_k$.
\end{thm}

% section free_abelian_groups (end)

% section simplices_and_simplicial_complexes (end)

% chapter lecture_15_feb_11th (end)

\chapter{Lecture 16 Feb 13th}%
\label{chp:lecture_16_feb_13th}
% chapter lecture_16_feb_13th

\section{Simplices and Simplicial Complexes}%
\label{sec:simplices_and_simplicial_complexes}
% section simplices_and_simplicial_complexes

We defined what the homology of a triangulated surface was, and we used these
triangles as the `building blocks' of any shape.

\begin{eg}
  Recall that a 2-face is as shown in \cref{fig:typical_2_face}.
  \begin{marginfigure}
    \centering
    \begin{tikzpicture}
      \draw (0, 1) -- (1, 0) -- (-1, 0) -- cycle;
    \end{tikzpicture}
    \caption{Typical 2-face}
    \label{fig:typical_2_face}
  \end{marginfigure}
  This is actually called a $2$-simplex.
\end{eg}

We did not give an explicit definition as to what a \hlnotea{face} is, but we
shall do that in this section.

\begin{defn}[Simplices]\index{Simplices}\index{Simplex}\label{defn:simplices}
  The \hlnoteb{$k$-simplex} is defined as a set
  \begin{equation*}
    \sigma^{(k)} = \left\{ v_0 + \sum_{j=1}^{k} t_j (v_j - v_0) : 0 \leq t_j
    \leq 1, \, \sum_{j=1}^{k} t_j = 1 \right\}.
  \end{equation*}
\end{defn}
\begin{marginfigure}
  \centering
  \includegraphics[width=\marginparwidth]{images/typical-2-simplicial-complex.png}
  \caption{A typical simplicial complex}\label{fig:a_typical_simplicial_complex}
\end{marginfigure}

\begin{defn}[Face]\index{Face}\label{defn:face}
  Let $\sigma$ be a $k$-simplex spanned by the vertices
  \begin{equation*}
    \mathcal{A} = \{ a_0, a_1, \ldots, a_k \}.
  \end{equation*}
  Any simplex spanned by a subset of $\mathcal{A}$ of size $l \leq k$ is called
  an \hlnoteb{$l$-face}.
\end{defn}
\begin{marginfigure}
  \centering
  \includegraphics[width=\marginparwidth]{images/faces-on-the-2-simplicial-complex.png}
  \caption{Faces on a simplicial complex}\label{fig:faces_on_a_simplicial_complex}
\end{marginfigure}

\begin{defn}[Simplicial Complex]\index{Simplicial Complex}\label{defn:simplicial_complex}
  A collection, $K$, of $k$-simplices is called a \hlnoteb{simplicial complex}
  if
  \begin{enumerate}
    \item every face of a simplex in $K$ is also in $K$; and
    \item the intersection of any two simplices in $K$ is a face of both of
      them.
  \end{enumerate}
\end{defn}

\begin{eg}
  The diagram in \cref{fig:non_example_for_a_simplicial_complex} is not a
  simplicial complex. For example, the edge $[a_3 \, a_4]$ is not an
  intersection of two $2$-faces.
  \begin{marginfigure}
    \centering
    \includegraphics[width=\marginparwidth]{images/non-example-simplicial-complex.png}
    \caption{Non-example for a simplicial complex}\label{fig:non_example_for_a_simplicial_complex}
  \end{marginfigure}
\end{eg}

\newthought{Recall} how it is common for us to represent a vertex as simply $a_0$ and an edge
as $[a_0 \, a_1]$. We may, for example, represent the edges in
\cref{fig:faces_on_a_simplicial_complex} as
\begin{equation*}
  [a_0 \, a_1], \, [a_1 \, a_2], \, [a_2 \, a_3], \, [a_3 \, a_0], \, [a_0 \,
  a_2], \, [a_3 \, a_1],
\end{equation*}
and the $2$-faces as
\begin{equation*}
  [a_0 \, a_1 \, a_2], \, [a_0 \, a_2 \, a_3], \, [a_0, \, a_3 \, a_1], \, [a_1
  \, a_2 \, a_3].
\end{equation*}
Note that the order of which we write the vertices in $[ \hdots ]$ indicates an
orientation. For instance,
\begin{itemize}
  \item $[a_0 \, a_1]$ is the edge that has $a_0$ as its starting point, and
    $a_1$ its ending point;
  \item $[a_0 \, a_1 \, a_2]$ is the $2$-face that is oriented anticlockwise.
\end{itemize}

\begin{note}\label{note:various_ways_to_represent_an_edge}
  In relation to how we write the above, for edges, we represent the opposite
  orientation by simply writing the vertices in reverse. E.g. the opposite of
  $[a_0 \, a_1]$ is $[a_1 \, a_0]$. We relate the two together by denoting one as
  the negative of the other:
  \begin{equation*}
    [a_0 \, a_1] = - [a_1 \, a_0].
  \end{equation*}
  This notation is consistent with the use of the
  \hyperref[defn:boundary_of_an_edge]{boundary operator} introduced earlier; in
  particular
  \begin{equation*}
    \partial ([a_0 \, a_1]) = a_1 - a_0 = - (a_0 - a_1) = - \partial ([a_1 \,
    a_0]).
  \end{equation*}
\end{note}

We may also do so for other simplices.

\begin{defn}[Boundary of Simplicies]\index{Boundary of Simplicies}\label{defn:boundary_of_simplicies}
  Let $[a_0 \hdots a_k]$ be a $k$-simplex. We define the boundary of $[a_0
  \hdots a_k]$ as
  \begin{equation*}
    \partial[a_0 \hdots a_k] = \sum_{j=0}^{k} (-1)^j [a_0 \hdots \hat{a_j}
    \hdots a_k],
  \end{equation*}
  where $\hat{a_j}$ indicates the removal of $a_j$ from the notation.
\end{defn}

\begin{eg}
  Let $\sigma = [a_0 \, a_1 \, a_2]$ be a $2$-face. The boundary of $\sigma$ is
  \begin{equation*}
    \partial \sigma = [a_1 \, a_2] - [a_0 \, a_2] + [a_0 \, a_1]
    = [a_1 \, a_2] + [a_2 \, a_0] + [a_0 \, a_1].
  \end{equation*}
  Notice that the last expression expresses the geometric boundary of $\sigma$,
  with an induced orientation.

  Also, notice that we may also write $\sigma$ as
  \begin{equation*}
    [a_1 \, a_2 \, a_0], \, [a_2 \, a_0 \, a_1].
  \end{equation*}
  In \cref{note:various_ways_to_represent_an_edge}, we learned that we can
  represent
  \begin{equation*}
    [a_0 \, a_1] = - [a_1 \, a_0].
  \end{equation*}
  We can do the same for $k$-simplices for any $k$; in particular,
  \begin{equation*}
    [a_0 \, \hdots \, a_k] = (\sgn \tau) [a_{\tau(0)} \hdots a_{\tau(k)}]
  \end{equation*}
  for $\tau \in S_k$, where $\sgn \tau$ is the
  \href{https://en.wikipedia.org/wiki/Parity_of_a_permutation}{sign} of $\tau$,
  and $S_k$ is the \hlnotea{permutation group on $\{1, \ldots, k\}$}.
\end{eg}

% section simplices_and_simplicial_complexes (end)

\section{Relative Homology}%
\label{sec:relative_homology}
% section relative_homology

\begin{defn}[Simplicial Subcomplex]\index{Simplicial Subcomplex}\label{defn:simplicial_subcomplex}
  Given a \hyperref[defn:simplicial_complex]{simplicial complex} $K$, a
  \hlnoteb{simplicial subcomplex} of $K$ is a subset of the simplicies in $K$
  that still forms a simplicial complex.
\end{defn}

Since each of the sets of $p$-chains form abelian groups, any of its subgroups
are normal, and in particular, if $K_0$ is a subcomplex of $K$, then we can take
\begin{equation*}
  \quotient{C_p(K)}{C_p(K_0)}.
\end{equation*}

\begin{defn}[Relative Chain]\index{Relative Chain}\label{defn:relative_chain}
  Given a simplicial complex $K$ and a subcomplex $K_0$,
\end{defn}

% section relative_homology (end)

% chapter lecture_16_feb_13th (end)

\chapter{Lecture 17 Feb 15th}%
\label{chp:lecture_17_feb_15th}
% chapter lecture_17_feb_15th

% chapter lecture_17_feb_15th (end)

\appendix

\chapter{Assignment Problems}%
\label{chp:assignment_problems}
% chapter assignment_problems

\section{Assignment \#1}%
\label{sec:assignment_1}
% section assignment_1

\begin{enumerate}
  \item
    \begin{enumerate}
      \item Find an explicit homeomorphism between the interval $(0, 1)$ and the real line
        $\mathbb{R}$.
      \item[(a')] Extend the last proof to show that the unit ball in $\mathbb{R}^n$ is
        homeomorphic to $\mathbb{R}^n$.
      \item Show that the circle, $S^1$, is not homeomorphic to any $S^m$ for
        $m \geq 2$.
      \item[(b')] Show that $S^1$ is not homeomorphic to $[0, 1)$.
      \item Prove that $[0, 1]$ is not homeomorphic to $\mathbb{R}$.
      \item Show that the interval $[0, 1)$ is not homeomorphic to the real line.
    \end{enumerate}

  \item Recall that the topologist's sine curse (similar to the topologist's circle)
    was defined as
    \begin{equation*}
      f(x) := \begin{cases}
        \sin \left( \frac{1}{x} \right) & x \in (0, 1] \\
        0                               & x = 0
      \end{cases}.
    \end{equation*}
    Show that the space given as the graph of this function, $\Gamma(f)$, is connected,
    but not path-connected or even locally connected.
  \item
    \begin{enumerate}
      \item Show that in a Hausdorff space, the limit of a convergent sequence is unique.
      \item Show that the subspace topology induced on a subset of a Hausdorff space is
        also Hausdorff.
    \end{enumerate}
  \item
    \begin{enumerate}
      \item Recall that the projection operator $\pi_1 : Y_1 \times Y_2 \to Y_1$ is defined
        by $\pi_1 : (y_1, y_2) \mapsto y_1$. Show that $\pi_1$, and similarly $\pi_2$, are
        continuous when $Y_1 \times Y_2$ is equipped with the product topology.
      \item Let $f : X \to Y_1 \times Y_2$ be a continuous map. Define $f_j : \pi_j \circ f$.
        Show that $f$ is continuous iff $f_1$ and $f_2$ are continuous.
      \item Let $g : X_1 \times X_2 \to Y$ be a function. We say that $g$ is continuous in
        each variable iff for any given $x_0$ and $y_0$, the functions $g(x_0, \cdot)$ and
        $g(\cdot, y_0)$ are continuous. Show that if $g$ is continuous, then it is continuous
        in each variable.
      \item Show that the converse of the last statement does not hold.
      \item Is the projection map $\pi_1$ an open map, a closed map, or neither?
    \end{enumerate}
  \item
    \begin{enumerate}
      \item Show that the image of a connected set is connected.
      \item Prove the intermediate value theorem for functions defined on the interval.
      \item Show that any continuous function $f : [0, 1] \to [0, 1]$ has a fixed point, that
        is a point $x$ such that $f(x) = x$.
    \end{enumerate}
  \item Consider the metric given by $\tilde{d}(x, y) := \arctan(\abs{ x- y })$. Determine
    if the topology induced is equivalent to the standard topology of $\mathbb{R}^n$. Further,
    determine if this metric satisfies the Heine-Borel theorem, i.e. is compactness of a subset
    equivalent to it being closed and bounded with respect to $\tilde{d}$.
\end{enumerate}

% section assignment_1 (end)

\newpage

\section{Assignment \#2}%
\label{sec:assignment_2}
% section assignment_2

\begin{enumerate}
  \item
    \begin{enumerate}
      \item Let $M$ be a manifold with or without boundary. Show that $\Int M$ is closed
        iff $\partial M = \emptyset$.
      \item Show that the boundary of an $n$-dimensional manifold is an $(n - 1)$-dimensional
        manifold \textit{without} boundary.
      \item Show that $[0, 1)$ is not homeomorphic to $(0, 1)$ and use that idea to prove
        that the invariance of boundary for $1$-dimensional manifolds.
      \item Prove the invariance of dimension for $1$-dimensional manifolds.
    \end{enumerate}
  \item
    \begin{enumerate}
      \item Show that the closed quadrant $M := \{ (x, y) \in \mathbb{R}^2 : x, y \geq 0 \}$
        is a manifold with boundary. Describe the coordinate charts explicitly.
      \item[(a')] Generalise what you proved in part (a) to
        $N := \{ (x_1, \ldots, x_n) \in \mathbb{R}^n : x_1, \ldots, x_n \geq 0 \}$,
      \item Let $M_1^m$ and $M_2^n$ be two topological manifolds of dimensions $m$ and $n$
        respectively. Take $N := M_1 \times M_2$ with its product topology. Show that $N$
        is an $(n + m)$-dimensional manifold whose boundary is
        $\partial N = M_1 \times \partial M_2 \cup \partial M_1 \times M_2$. Conclude, in
        particuar, that $N$ has no boundary iff $M_1 $ and $M_2$ have empty boundary.
    \end{enumerate}
  \item Let $P$ be an arbitrary $n$-gon, that is to say, a union of finite number of line
    segments such that each line end point is shared by exatly two segments. Show that $P$
    is homeomorphic to $S^1$. (You might find the glueing lemma stated as Lemma 3.23 of J.
    Lee's book useful.)
  \item Let $f : M_1 \to M_2$ be a homeomorphism between two manifolds with a boundary.
    Show that under $f$ the boundary of $M_1$ is sent to the boundary of $M_2$. Conclude
    that in such a case we must have $\partial M_1 \simeq \partial M_2$.
  \item Let $\mathbb{T}^2$ be the standard torus obtained by taking the quotient
    $\mathbb{R}^2 / (\mathbb{Z} \oplus \mathbb{Z})$ with the projection map
    $\Pi : \mathbb{R}^2 \to \mathbb{T}^2$. Define $L$ to be the line $y = \sqrt{2}x$ in
    $\mathbb{R}^2$.
    \begin{enumerate}
      \item Determine if the image of $L$ under this projection, $\Pi(L)$, is a
        manifold or not.
      \item Show that $\Pi(L)$ is the image of $\mathbb{R}$ under a bijective continuous
        map, but it is not homeomorphic to $\mathbb{R}$. Conclude that $\Pi \restriction_{L}$
        is not an embedding of $L$.
    \end{enumerate}
  \item Recall that we showed in class how to obtain the cylinder as the quotient of the
    plane by the action of $\mathbb{Z}$. Define a group action of $G$ on $\mathbb{R}^2$
    such that $\mathbb{R}^2 / G$ will be homeomorphic to the infinite M\"{o}bius band.
\end{enumerate}

% section assignment_2 (end)

\newpage

\section{Assignment \#3}%
\label{sec:assignment_3}
% section assignment_3

\begin{enumerate}
	\item Find an injective open map between topological spaces which is not continuous.
  \item Consider the M\"{o}bius band with boundary, $\mathfrak{M}$.
    \begin{enumerate}
      \item Find an explicit embedding $f : \mathfrak{M} \to \mathbb{R}^3$ using elementary
        functions.
      \item Generalise that embedding to the case of the space obtained by $k$ times twisting
        and then glueing the unit cell $[0, 1] \times [0, 1]$ along $\{ 0 \} \times [0, 1]$ and
        $\{ 1 \} \times [0, 1]$.
      \item Conclude that all bands with an even number of twists are homeomorphic to the cylinder
        $S^1 \times [0, 1]$ and the ones with an odd number of twists are homeomorphic to
        $\mathfrak{M}$.
    \end{enumerate}
  \item Consider the action of $\mathbb{Z}$ on $M := \mathbb{R}^n \setminus \{0\}$
    defined by $k \cdot (x_1, \ldots, x_n) := (2^k x_1, \ldots, 2^k x_n)$.
    \begin{itemize}
      \item Show that for any element $k \in \mathbb{Z}$, the action on $M$ by $k$ has no fixed
        points. (Recall that for an element $g \in G$ acting on a space $X$, we say that $x \in
        X$ is a fixed point provided that $g \cdot x = x$.)
      \item Can you identify the quotient space $M / \mathbb{Z}$? (It might help to start from
        lower dimensions.)
    \end{itemize}
  \item (Embedding the projective plane) Recall how we defined the projective plane
    $\mathbb{P}^2$ as the quotient $S^2 / ((x, y, z) \sim -(x, y, z))$. This can be generalized to
    the case of $\mathbb{P}^n$, which is defined as the quotient of $S^n \subset \mathbb{R}^{n + 1}$
    by the antipodal map $w \in \mathbb{R}^{n + 1} \mapsto - w$.
    \begin{enumerate}
      \item Show that any $\mathbb{P}^n$ embeds in $\mathbb{P}^{n + 1}$. (Hint: It can be done in
        two lines.)
      \item Consider the map $F : \mathbb{R}^3 \to \mathbb{R}^4$ defined as $(x, y, z) \mapsto
        (xy, yz, xz, x^2 - y^2)$. Show that the map $f := F \restriction_{S^2}$ has the property
        that $f(x, y, z) = f(-x, -y, -z)$. Use this observation to show that this map descends
        to a continuous map $\tilde{f} : \mathbb{P}^2 \to \mathbb{R}^4$.
      \item Show that $\tilde{f}$ is injective. Notice that $\mathbb{P}^2$ is compact, hence, any
        continuous map is a closed map. (Indeed, we stated this as a general lemma: any continuous
        injection from a compact space is an embedding.) Combine these two observations to prove
        that $\tilde{f}$ gives an embedding of $\mathbb{P}^2$ into $\mathbb{R}^4$.
    \end{enumerate}
  \item Recall that we found the homology groups of sphere $S^2$ by finding the
    homology groups of a (boundary of) a tetrahedron (cf. \cref{fig:_3_simplex_tetradedron}).
    We can see this tetrahedron as the manifold boundary of a $3$-simplex.
    \begin{figure}[ht]
      \centering
      \begin{tikzpicture}
        \draw (0,1,0) -- (-1,0,-1) -- (0,0,1) -- (0,1,0) -- (1,0,0) -- (0,0,1);
        \draw[dashed] (-1, 0, -1) -- (1, 0, 0);
      \end{tikzpicture}
      \caption{$3$-simplex, tetradedron}
      \label{fig:_3_simplex_tetradedron}
    \end{figure}
    Let $K_n$ denote the simplicial complex formed by the boundary of an $(n + 1)$-simplex.
    Generalise what we did in class in case of the tetrahedron to show that we have
    \begin{equation*}
      H_0(K_n) = \mathbb{Z}, H_1(K_n) = 0, \ldots, H_{n - 1}(K_n) = 0, H_n(K_n) = \mathbb{Z}.
    \end{equation*}
    Let us assume the result that says homology is invariant under homeomorphism. Conclude that
    we have for the $n$-dimensional sphere, $S^n$:
    \begin{equation}
      H_0(S^n) = \mathbb{Z}, H_1(S^n) = 0, \ldots, H_{n - 1}(S^n) = 0, H_n(S^n) = \mathbb{Z}.
    \end{equation}
  \item Verify that the triangulated surface in the figure below is homeomorphic to the
    M\"{o}bius strip and find its homology groups.
    \begin{figure}[ht]
      \centering
      \begin{tikzpicture}
        \draw (-1, 1) -- (-1, 0) -- (1, 0) -- (1, 1) -- cycle;
        \draw (-1, 0) -- (0, 1) -- (1, 0);
        \draw (0, 0) -- (0, 1);
        \node[above,left] at (-1, 1) {$\bm{a}$};
        \node[below,left] at (-1, 0) {$\bm{b}$};
        \node[below,right] at (1, 0) {$\bm{a}$};
        \node[above,right] at (1, 1) {$\bm{b}$};
        \node[above] at (0, 1) {$\bm{c}$};
        \node[below] at (0, 0) {$\bm{d}$};
      \end{tikzpicture}
      \caption{M\"{o}bius band}
      \label{fig:ass3_mobius_band}
    \end{figure}
  \item Recall that a holomorphic map that is not constant is an open map. More generally,
    a meromorphic function can be seen as an open map when the target space is taken to be the
    extended plane $\mathbb{C} \cup \{ \infty \}$. This is the content of the open mapping theorem
    in complex analysis. Conoclude that a non-constant holomorphic (meromorphic) map is a quotient
    map. In particular, study the case of the following function wherein $k$ is an integer.
    \begin{equation*}
      \begin{cases}
        f_k : \mathbb{C} \cup \{ \infty \} \to \mathbb{C} \cup \{ \infty \} \\
        z \mapsto z^k
      \end{cases}
    \end{equation*}
    Show that $f_k$ is holomorphic to the extended complex plane and may, hence, be seen as a
    quotient map $S^2 \to S^2$. Identify the various types of fibres over various points of the
    target.
    \begin{enumerate}
      \item Consider the action on $\mathbb{C}$ of $\mathbb{Z}_m$ given by multiplication by
        $\zeta_m$, the $m$-th root of unity. Describe the corresponding action of $\mathbb{Z}_m$
        on the sphere $S^2$. Show that under this action $\mathbb{Z}_m$ on $S^2$, we have
        $S^2 / \mathbb{Z}_m \simeq S^2$. \\
        (\textbf{Remark}: This equivalence is only at the level of homeomorphism and does not
        hold at the level of diffeomorphism.)
      \item Let $\Lambda$ be a lattice in $\mathbb{R}^2 = \mathbb{C}$. Show that a function can be
        seen as a continuous function on the torus $\mathbb{C} / \Lambda$ iff it is
        $\Lambda$-periodic.
      \item[(b')] (Optional) Consider the following function defined on $\mathbb{C}$. Show that it
        is doubly periodic where the lattice $\Lambda$ is generated by $1$ and $i$.
        \begin{equation*}
          \wp(\xi) := \frac{1}{\xi^2} +
            \sum_{\mu^2 + \nu^2 \neq 0} \left( \frac{1}{(\xi + m + n i)^2}
              - \frac{1}{(m + ni)^2} \right)
        \end{equation*}
        (You can do so by showing that the derivative $\frac{\dif{\wp}}{\dif{z}}$ is a doubly
        periodic function. Indeed, it is easy to show that the derivative converges away from
        the poles. Further, $\wp$ is an even function. This, along with the double periodicity
        of the derivative, gives periodicity of $\wp$.) This is an example of a Weierstrass
        $\wp$-function.
      \item Use the previous observations to conclude that this function descends to a continuous
        function on the torus $T$ obtained from $[0, 1] \times [0, 1] \subset \mathbb{R}^2$.
        Indeed, more is true. We shall not go into the details in full generality, but since $\wp$
        is meromorphic on the complex plane, it descends to a meromorphic function on $T$. You
        can verify meromorphicity by studying the function along the boundaries of the torus.
      \item Show that $\wp$ provides a meromorphic function with two poles on $\mathbb{T}^2$. In
        particular, it is a quotient map from the torus $\mathbb{T}^2$ to the sphere $S^2$.
    \end{enumerate}
\end{enumerate}

% section assignment_3 (end)

% chapter assignment_problems (end)

\backmatter

\pagestyle{plain}

\nobibliography*
\bibliography{references}

\printindex

\end{document}
% vim: tw=80