I am reading Lee's Introduction to Smooth Manifolds and I have some problems with definition of Grassmannian manifold given in Example 1.24, [p.22](http://books.google.com/books?id=eqfgZtjQceYC&pg=PA22). I'll write the details below.

My question is:

* Why is the set $\varphi(U\_Q \cap U\_{Q'}) \subset L(P, Q)$ (i.e., the set of all
$A \in L(P, Q)$ whose graphs intersect both $Q$ and $Q'$ trivially) an open set?
(Which topology is used on $L(P,Q)$. Should I take the standard topology from $\mathbb R^{k(n-k)}$ and an identification of this space with $L(P,Q)$?)

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Here is the relevant part from the book:

> 
> **Example 1.24 (Grassmann Manifolds)**. Let $V$ be an $n$-dimensional
>  real vector space. For any integer $0 \le k \le n$, we let $G\_k(V)$ denote the set
>  of all $k$-dimensional linear subspaces of $V$ . We will show that $G\_k(V)$ can be
>  naturally given the structure of a smooth manifold of dimension $k(n-k)$.
>  The construction is somewhat more involved than the ones we have done
>  so far, but the basic idea is just to use linear algebra to construct charts for
>  $G\_k(V)$ and then use the smooth manifold construction lemma (Lemma 1.23).
> 
> 
> Let $P$ and $Q$ be any complementary subspaces of $V$ of dimensions $k$ and
>  $(n-k)$, respectively, so that $V$ decomposes as a direct sum: $V = P \oplus Q$.
>  The graph of any linear map $A \colon P \to Q$ is a $k$-dimensional subspace $\Gamma(A) \subset V$,
>  defined by
>  $$\Gamma(A)=\{x+Ax: x\in P\}.$$
>  Any such subspace has the property that its intersection with $Q$ is the zero
>  subspace. Conversely, any subspace with this property is easily seen to be
>  the graph of a unique linear map $A\colon P \to Q$.
> 
> 
> Let $L(P, Q)$ denote the vector space of linear maps from $P$ to $Q$, and
>  let $U\_Q$ denote the subset of $G\_k(V)$ consisting of $k$-dimensional subspaces
>  whose intersection with $Q$ is trivial. Define a map $\psi \colon L(P, Q) \to U\_Q$ by
>  $$\psi(A)=\Gamma(A).$$
>  The discussion above shows that $\psi$ is a bijection. Let $\varphi = \psi^{-1} \colon U\_Q \to
> L(P, Q)$. By choosing bases for $P$ and $Q$, we can identify $L(P, Q)$ with
>  $M((n-k)\times k, \mathbb R)$ and hence with $\mathbb R^{k(n-k)}$, and thus we can think of $(U\_Q,\varphi)$
>  as a coordinate chart. Since the image of each chart is all of $L(P, Q)$, condition (i) of Lemma 1.23 is clearly satisfied.
> 
> 
> Now let $(P', Q')$ be any other such pair of subspaces, and let $\psi'$, $\varphi'$ be
>  the corresponding maps. The set $\varphi(U\_Q \cap U\_{Q'} ) \subset L(P, Q)$ consists of all
>  $A \in L(P, Q)$ whose graphs intersect both $Q$ and $Q'$ trivially, **which is easily
>  seen to be an open set,** so (ii) holds.
> 
> 
>