End of Hom-profunctor in Grp

Question

What is the end of the $\operatorname{Hom}$ profunctor in the category of groups?

$$\int_{X\in\mathsf{Grp}} \operatorname{Hom}(X, X)$$

Preface: if we examine ends of $\operatorname{Hom}(-, -)$ in other categories, in $\mathsf{Set}$ we get the identity function, or rather a mapping $S \mapsto \{\operatorname{id}_S\}$.

In $\mathsf{Vect}_K$ the end is $\{(k \cdot -) : k \in K\}$ - the set of uniformly scaling linear maps, and if we take the end of the internal $\operatorname{hom} : \mathsf{Vect}_K^{op} \times \mathsf{Vect}_K \to \mathsf{Vect}_K$, we get the vector space of those maps (isomorphic to $K$).

Intuitively, the end of $\operatorname{Hom}$ seems to be the set of morphisms (or rather morphism families parameterised by the domain object) that can be applied to any object, $\{x \mapsto x\}$ in $\mathsf{Set}$ and $\{(x \mapsto k\cdot x) | k \in K\}$ in $\mathsf{Vect}_K$.

Continuing the line of thought the end of $\operatorname{Hom}$ in $\mathsf{FinGrp}$ and likely $\mathsf{Grp}$ too, is $\{(x \mapsto x^i) | i \in \mathbb{Z}\}$.

The proof that $G \mapsto \{(x \mapsto x^i) : G \to G | i \in \mathbb{Z}\}$ is a wedge of $\operatorname{Hom}(-, -)$ is simple enough, for any group homomorphism $f : G \to H$, $f(x^i) = f(x)^i$ and hence pre- and post-compositions of a hom-set with $-^i$ are necessarily equal.

However I have no proof that this wedge is universal.

Answer 1

You're either approaching, or already made without saying so explicitly, the realization that the end of the hom bifunctor is the set of natural endomorphisms of the identity functor. This follows directly from the definition, once you unravel it. This immediately gives us the first two results you list: since a singleton represents the identity endofunctor of $\mathbf{Set}$, its single endomorphism represents the end of the hom bifunctor, and analogously in vector spaces. Since groups aren't enriched over themselves, we can't make quite the same argument to calculate $\int_G\mathrm{Gp}(G,G)$ as $\mathrm{Gp}(\mathbb{Z},\mathbb{Z})$. However, there's a nice trick we can use: groups have a faithful forgetful functor $U$, and so a natural endormorphism of the identity is just a natural endomorphism of the forgetful functor whose components happen to be group homomorphisms. But the forgetful functor is represented by $\mathbb{Z}$, so its endomorphisms are just $\mathrm{Gp}(\mathbb{Z},\mathbb{Z})$, each element of which corresponds to the natural endomorphism $g\mapsto g^n$ of $U$. However, none of these endomorphisms of $U$ are group homomorphisms for nonabelian groups $G$, except when $n=1$ or $n=0$! Thus there are only two elements in the end of the hom bifunctor for groups, corresponding to the identity natural transformation and to that which maps each group to its identity element.

The object we've just computed is often known as the center of the category, as it's guaranteed to be a commutative monoid by the Eckmann-Hilton argument. Note that, indeed, all of our computations so far are commutative monoids, mostly rather boring ones.