# Can the map sending a presentation to its group be considered as a functor?

It is well-known that the functor $$Grp \to Set$$ sending a group $$G$$ to its underlying set $$UG$$ has a left adjoint, the functor $$Set\to Grp$$ sending a set $$X$$ to the free group $$FX$$.

I wonder whether one can consider the following modification of the functor $$F$$: to each presentation $$\langle S\mid R\rangle$$ we assign the group $$FS/N$$, where $$N$$ is the normal subgroup induced by $$R$$.

1. Is this a functor?
2. For this to be true, there need to be a category of presentations. Is there such a category?
3. Does the functor $$\langle S\mid R\rangle\mapsto FS/N$$ have a right adjoint (maybe similar to $$U$$)?

Of course, my questions are not precise, because I haven't specified what the morphisms between presentations are. The question is if one can do all this in a sensible way.

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• This (not entirely serious but nonetheless valid) paper constructs such a functor using some high-powered machinery. – Zhen Lin Nov 20 at 22:45

Yes, yes, yes, to all three questions.
And it can be done very generally and very nicely for universal algebra $$\$$ [or even for first order structures].

Let's fix an algebraic signature consisting of named finitary operation symbols, and a variety $$\mathcal{Alg}$$ of algebras of that signature that we're working on. (In your question, the groups.)
As in your question, we use $$F:\mathcal{Set}\to\mathcal{Alg}$$ for the free functor and $$U:\mathcal{Alg}\to\mathcal{Set}$$ for the underlying set functor.

A presentation $$\langle A,E\rangle$$ is a pair where $$A$$ is a set and $$E$$ is a subset of equations, i.e. pairs of terms (in other words $$E\subseteq F(A)^2$$) $$\$$ [or, $$E$$ is a set of atomic formulas].
In the case of groups, an equation can be arranged on one side, leaving the identity on the other side, so then $$E\subseteq F(A)$$ is enough, as learned.

We can define a morphism of presentations $$\langle A,E\rangle\to\langle B,\Sigma\rangle$$ to simply be a function $$f:A\to B$$ such that $$(\tau,\sigma)\in E\ \implies\ (Ff(\tau),\,Ff(\sigma))\in\Sigma\,.$$

By an interpretation of the presentation $$\langle A,E\rangle$$ in an algebraic structure $$\mathfrak B$$ we mean a function $$h:A\to U\mathfrak B$$ such that all elements of $$E$$ are evaluated equal $$\,$$ [all atomic formulas of $$E$$ are true], i.e. for every $$(\tau,\sigma)\in E$$ we have $$\varepsilon_{\mathfrak B}\left( Fh(\tau)\right) \ =\ \varepsilon_{\mathfrak B}\left( Fh(\sigma)\right)\ \text{ that is, }\ (\tau,\sigma)\in\ker(\varepsilon_{\mathfrak B}\circ Fh)$$ where $$\varepsilon:FU\to 1_{\mathcal{Alg}}$$ is the counit of the free-forgetful adjunction.

Observe that we can compose morphisms of presentations with interpretations as well as with homomorphisms, in an associative way, since all are just functions.
So this actually connects up the category of presentations $$\mathcal{Pres}$$ with the category of algebras $$\mathcal{Alg}$$ by (so called hetero-)morphisms between them, yielding to a bigger category, containing both $$\mathcal{Pres}$$ and $$\mathcal{Alg}$$ as a full subcategory.
Such an animal is called a profunctor, and it is determined by the restriction of its hom functor to the heteromorphisms, i.e. of a functor of the form $$\mathcal{Pres}^{op}\times\mathcal{Alg}\to\mathcal{Set}$$.

Now, every presentation $$\langle A,E\rangle$$ has a reflection in $$\mathcal{Alg}$$, namely $$FA/(E)$$ where $$(E)$$ is the congruence relation generated by $$E$$.
This defines the free functor on presentations.

Conversely, every algebraic structure $$\mathfrak B$$ has a coreflection in $$\mathcal{Pres}$$, namely $$\langle U\mathfrak B,\Delta_{\mathfrak B}\rangle$$ where $$\Delta_{\mathfrak B}$$ is the set of all equations that can be written on letters of $$U\mathfrak B$$ and hold in $$\mathfrak B$$ $$\,$$ [the positive diagram of $$\mathfrak B$$].
• Note that morphisms of presentations can indeed be defined more ways, but the one in the answer works clearest, at least for getting the right adjoint of the free functor. One can alternatively consider a poset by inclusion (i.e. there's a unique arrow $\langle A,E\rangle\to\langle B,\Sigma\rangle$ if $A\subseteq B$ and $E\subseteq\Sigma$). Or, we can regard a function $f:A\to B$ as a morphism $\langle A,E\rangle\to\langle B,\Sigma\rangle$ if the ($f$-transferred) equations in $E$ are consequences of those of $\Sigma$, optionally adding also a proof of each as part of the morphism. – Berci Nov 21 at 20:25
Giving a presentation $$\langle S \mid R \rangle$$ of a group amounts to describing it as the cokernel of a map $$F(R) \to F(S)$$ between free groups. There is a category $$C$$ whose objects are such maps and whose morphisms $$(F(R_1) \to F(S_1)) \to (F(R_2) \to F(S_2))$$ are maps $$f : S_1 \to S_2$$ such that the induced map $$F(f) : F(S_1) \to F(S_2)$$ sends every relation in $$R_1$$ to an element in the image of $$F(R_2)$$. Taking the cokernel to get the group is a functor on this category, which has a right adjoint sending a group $$G$$ to the map $$1 \to F(G)$$.
$$\text{Hom}_{\text{Grp}}(\langle S \mid R \rangle, G) \cong \text{Hom}_C(F(R) \to F(S), 1 \to F(G))$$
which follows since both sides describe the set of functions $$f : S \to G$$ such that every element of $$F(R)$$ is sent to the identity.