# Group Presentation - Categorical Interpretation

In this post, user Martin Brandenburg wrote

$$\langle x_1,x_2,\dotsc \mid R_1,R_2,\dotsc \rangle$$ is defined to be a group satisfying the following universal property:

The group has elements $x_1,x_2,\dotsc$ satisfying the relations $R_1(x),R_2(x),\dotsc$, such that for every other group $G$ and every other sequence of elements $g_1,g_2,\dotsc$ in $G$ satisfying the relations $R_1(g), R_2(g), \dotsc$ in $G$ there is a unique homomorphism $\phi : \langle x_1,x_2,\dotsc \mid R_1,R_2,\dotsc \rangle \to G$ which maps $\phi(x_i)=g_i$.

In category-theoretic language, this means that $\langle x_1,x_2,\dotsc \mid R_1,R_2,\dotsc \rangle$ represents the functor $\mathsf{Grp} \to \mathsf{Set}$ which maps $G$ to the set of all families of elements $(g_1,g_2,\dotsc)$ in $G$ satisfying the relations $R_1(g),R_2(g),\dotsc$ in $G$.

Problem: I don't see how one comes up with the functor above from the universal property formulation of group, i.e. how did it make the deduction "In category-theoretic language, this means..." ?

Sure we can define a functor (as my attempt below); but how did we even think of that? Sorry if this is obvious, I am quite new to Category Theory.

(Is this right?): This functor $F: \mathsf{Grp} \rightarrow \mathsf{Set}$, satisfies $F(A) := A'$, where $A'$ is as described in the OP. Now let $f:A \rightarrow B$ be a homomorphism. If $\{ x_1, \ldots, x_n, \ldots \}$ satisfies relations, then $\{ f(x_1, \ldots, f(x_n), \ldots \}$ satisfies relations in $B$ ($f$ is homomorphism). Hence, define $F(f)(\{a_i \}) := \{f(a_i) \}$ for all $\{a_i \} \in A'$, and $F$ satisfies axioms of functor.

• Your note is right in the sense of defining the functor sending $G$ to the family of elements in $G$ satisfying the relations. But to show that your functor $F$ is represented by the group presentation $P$ you have to come up with a bijection between the set $Hom(P, G)$ and the set $F(G)$ that is natural in $G$. Apr 4, 2017 at 7:42

If you haven't appreciated it yet, start really caring about continuity properties and whether things can be represented as (co)limits. I also highly recommend getting as comfortable with the notion of representability as you can. Also, it's very useful to know what limits and colimits look like in $\mathbf{Set}$. As a general tip, whenever you hear "given ... there exists a unique ...", you should think to name that relation with a function from the givens, i.e. to Skolemize.

So, by definition, the notation $\langle x_1, x_2,\dotsc\mid R_1, R_2, \dotsc\rangle$ means calculate the quotient of the free group generated by the set $X=\{x_1,x_2,\dotsc\}$ by the equivalence relation generated by the relations $R_i$. We say the free group functor, $F : \mathbf{Set}\to\mathbf{Grp}$, is left adjoint to the obvious underlying set functor $U : \mathbf{Grp}\to\mathbf{Set}$. The quotient is, in this case, a coequalizer which can be represented by arrows $R_i : \mathbb{Z}\to FX$ all being coequalized with constantly identity map, which I'll write $e : \mathbb{Z}\to FX$. Incidentally, $\mathbb{Z}\cong F1$ which is why it is used. Here $1(=\{*\})$ is the terminal object in $\mathbf{Set}$, i.e. a singleton set (for which I've chosen to write the meaningless element as $*$).

Next, some generalities. $F\dashv U$ means $\lceil -\rceil :\mathbf{Grp}(FX,G)\cong\mathbf{Set}(X,UG) : \lfloor - \rfloor$ natural in $X$ and $G$. By this I mean $\lceil - \rceil : \mathbf{Grp}(FX,G)\to\mathbf{Set}(X,UG)$ and $\lfloor - \rfloor$ is its inverse going the opposite direction. We have $\lceil \varphi\circ\psi \rceil = U\varphi\circ\lceil \psi\rceil$ by naturality. I'll write $FX/{\sim}$ for the group we want to create, $\sim$ indicating the coequalizer. We want to explicitly spell out the universal property already inherent here. That is, we want an explicit articulation of the representable functor $\mathbf{Grp}(FX/{\sim},-)$ in terms of it's constituents. Now, for any hom-functor, it's continuous in both arguments, but since it's contravariant in the first argument, it "looks" like it takes colimits to limits. That is, we have $\text{Hom}(\mathsf{Colim}D,Y)\cong\mathsf{Lim}(I\mapsto\text{Hom}(DI,Y))$ natural in $Y$ (and for completeness $\text{Hom}(X,\mathsf{Lim}D)\cong\mathsf{Lim}(I\mapsto\text{Hom}(X,DI))$). Note that this turns colimits and limits in a category into limits in $\mathbf{Set}$. Limits in $\mathbf{Set}$ are calculated as equationally defined subsets of products. You'll see how this works out for equalizers momentarily. The corollary to the Yoneda lemma — that the Yoneda embedding is full and faithful — is what let's us move between $X \cong Y$ and $\text{Hom}(X,-)\cong\text{Hom}(Y,-)$ in either direction. Most of the time results will be proven by showing a (natural) isomorphism of hom-functors and the step using Yoneda to reduce that to an isomorphism of objects will be tacit.

Now to calculate. First, as a convenience, a lemma:\begin{align} \mathbf{Grp}(\mathbb{Z},G) & \cong \mathbf{Grp}(F1,G) \\ & \cong \mathbf{Set}(1,UG) \\ & \cong UG \end{align} This states that a group homomorphism from $\mathbb{Z}\to G$ is the same thing as an element of $UG$. I'll write $\bar R_i (= \lceil R_i \rceil(*))$ and $\bar e (=\lceil e\rceil(*))$ for the corresponding elements of $UFX$. Finally, I'll write $\varepsilon$ for the identity element of $G$.

\begin{align} \mathbf{Grp}(FX/{\sim},G) & \cong \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.\varphi \circ R_i = \varphi \circ e \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.\lceil\varphi \circ R_i\rceil = \lceil\varphi \circ e\rceil \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.U\varphi \circ \lceil R_i\rceil = U\varphi \circ \lceil e\rceil \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.\forall t\in 1.(U\varphi)(\lceil R_i\rceil(t)) = (U\varphi)(\lceil e\rceil(t)) \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.(U\varphi)(\lceil R_i\rceil(*)) = (U\varphi)(\lceil e\rceil(*)) \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.(U\varphi)(\bar R_i) = (U\varphi)(\bar e) \} \\ & = \{\varphi\in\mathbf{Grp}(FX,G)\mid \forall i.(U\varphi)(\bar R_i) = \varepsilon \} \\ & \cong \{g\in\mathbf{Set}(X,UG)\mid \forall i.(U\lfloor g\rfloor)(\bar R_i) = \varepsilon \} \end{align}

Most of the above is actually set theoretic reasoning not categorical reasoning which is part of why representability is convenient, as you are already likely to be familiar with set theoretic reasoning. I've also been very detailed in the above, taking small steps. Normally, I'd abbreviate most of the equational reasoning in the middle. While I've written in $X$ and $G$, this is all actually happening at the level of (natural) isomorphisms of hom-functors. The first isomorphism though, is continuity of the hom-functor in its first argument and so the right hand side is the explicit formulation of an equalizer in $\mathbf{Set}$, namely an equationally defined subset. Specifically, this is the equalizer of the set functions $\mathbf{Grp}(R_i,G):\mathbf{Grp}(FX,G)\to\mathbf{Grp}(\mathbb{Z},G)$ for each $i$ and similarly for $e$. $\mathbf{Grp}(R_i,G)(\varphi) = \varphi\circ R_i$ and similarly for $e$. So we could rewrite the first equation as $\mathbf{Grp}(R_i,G)(\varphi)=\mathbf{Grp}(e,G)(\varphi)$ which would more closely match the equalizer construction. Less noisily, the equalizer of the pair $f, g : X \to Y$ in $\mathbf{Set}$ is $\{ x\in X\mid f(x) = g(x)\}$.

Reading out the final natural isomorphism, it says, if you have a function $g$ going from the set of generators $X$ to elements of the (underlying set of the) group $G$, this can be lifted to a group homomorphism $\lfloor g\rfloor : FX \to G$, and if $\lfloor g \rfloor(\bar R_i) = \varepsilon$ for each $\bar R_i$, then $\lfloor g \rfloor$ descends to a homomorphism from $FX/{\sim}$ to $G$. The $\bar R_i$ are words on the alphabet $X$ and, by example, $\lfloor g \rfloor$ is $$\lfloor g \rfloor(x_1x_2x_1^{-1}x_2^{-1})=g(x_1)\cdot g(x_2)\cdot g(x_1)^{-1}\cdot g(x_2)^{-1}$$ where $\cdot$ is the multiplication in $G$. The concrete description of $FX$ does not arise from the work above but instead is part of the proof that the left adjoint $F$ exists.

I imagine the above will not be easy to follow the first time around due to the amount of detail and brisk pace. I did want to provide a worked example of going from "abstract nonsense" to concrete, explicit definitions. A lot of categorical work is highly constructive, but one typically doesn't spell out the details (because they're usually not that important). Nevertheless, even highly abstract category theory can usually be expanded into concrete set theoretic or type theoretic constructions.

• The importance of Skolemisation (in the categorical setting) cannot be overstated! Apr 4, 2017 at 13:30
• Perhaps expand on the type of $R_i$; are they "relations" or "functions" or what. Making clear what mathematical creature they are may make it clear how we jump to: $\mathbf{Grp}(R_i,G):\mathbf{Grp}(FX,G)\to\mathbf{Grp}(\mathbb{Z},G)$. Indeed, this latter item suggests, due to Yoneda, that $R_i : F\, X \to \mathbb{Z}$, and is that what is intended? If so, perhaps clarify and elaborate, as the asker claims to be "new" and you state you want to go from abstract to concrete ;) Apr 4, 2017 at 13:44
• @MusaAl-hassy $\mathbf{Grp}(-,G)$ is contravariant and $R_i$ is explicitly stated as being $\mathbb{Z}\to FX$. I definitely am ambiguously referring to $R_i$ as "relations" at the beginning. "Relator" would be a better word. I'll probably expand how the coequalizer leads to the quotient and clean up the wording in that section later. Arguably, the $\bar R_i$ are "really" what the relators are and coding them up as group homomorphisms is just so I can present the quotient as a coequalizer in $\mathbf{Grp}$. Apr 4, 2017 at 14:13
• Thanks a lot Derek, I believe I still have a lot to learn. i.e. limits, colimits, and adjoints etc... I will come back and review your post once I have better understanding :) Apr 4, 2017 at 14:58
• @CWL You may find my blog article interesting, or at least the references it cites, which I recommend, though "Basic Concepts of Enriched Category Theory" definitely isn't aimed at beginners. Depending on your background, you may also find this article interesting depending on your background, though it's definitely... non-traditional. Apr 5, 2017 at 3:52
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