How do I determine all the endomorphisms of the identity functor $\text{id}_{\text {Grp} }: \text {Grp}\rightarrow \text {Grp}$?

Question

I am trying to determine all the endomorphisms of the identity functor $\text{id}_{\text {Grp} }: \text {Grp}\rightarrow \text {Grp}$

I know :

-that the forgetful functor $F:\text {Grp}\rightarrow \text {Set}$ is representable

-The representing object is $\mathbb Z$

-Using the Yoneda lemma:$$\mathrm{Hom}(F,F) \cong \mathrm{Hom}(\mathrm{Hom}(\mathbb{Z},-),F) \cong F(\mathbb{Z}),$$

How do I use these facts to arrive to the thesis?

Answer 1

Note that the Yoneda Lemma only applies when talking about $\rm Set$-valued functors. However, the identity functor on $\rm Grp$ is evidently not $\rm Set $-valued.

In this particular case, a careful analysis of the objects involved yields the result in an elementary way. We start by unpacking what an endomorphism of the identity functor $\operatorname{id}_{\rm Grp}\colon{\rm Grp}\to{\rm Grp}$ actually is. By definition, this is a collection $$ (\alpha_G\colon G\to G)_{G\in{\rm Grp}} $$ of group homomorphisms. Among these groups is the group $G=\mathbb Z$. This group has two important properties:

• $\operatorname{Hom}_{\rm Grp}(\mathbb Z,\mathbb Z)\cong\mathbb Z$ via $(\varphi\colon\mathbb Z\to\mathbb Z)\mapsto\varphi(1)$
• $\operatorname{Hom}_{\rm Grp}(\mathbb Z,G)\cong\operatorname{Hom}_{\rm Set}(\{\bullet\},G)$ via $(f\colon\mathbb Z\to G)\mapsto(\{\bullet\}\to UG,\,\bullet\mapsto\varphi(1))$, i.e. group homomorphisms from $\mathbb Z$ into $G$ are in one-to-one correspondence with elements of $g$ (this is representability of the forgetful functor written slightly differently)

This shows immediately, that there are $n$ distinct choices for the map $\alpha_{\mathbb Z}$, uniquely determined by the image of $1$. Fix now one such $n$, i.e. suppose that $\alpha_{\mathbb Z}(1)=n$. If we now let $g\in G$, we can consider the unique group homomorphism $\varphi_g\colon\mathbb Z\to G$ mapping $1$ to $g$. The associated naturality square reads $\require{AMScd}$ \begin{CD} \mathbb Z @>{\alpha_{\mathbb Z}}>> \mathbb Z \\ @V{\varphi_G}VV @VV{\varphi_G}V \\ G @>>{\alpha_G}> G \end{CD} and we deduce that $$ \alpha_G(g) =\alpha_G(\varphi_G(1)) =\varphi_G(\alpha_{\mathbb Z}(1)) =\varphi_G(n) =g^n\,. $$ Thus, if we fix $n\in\mathbb Z$ so that $\alpha_{\mathbb Z}(1)=n$, then all $\alpha_G\colon G\to G$ are uniquely determined as $g\mapsto g^n$. However, letting $G=F_2$ be the free group in two generators, shows that only $n=0,1$ produce group homomorphisms.

Thus, there are exactly two endomorphisms of the identity functor, one with all components the trivial group homomorphism and one with all components identities.

In the same way one can classify endomorphisms of $\operatorname{id}_{\rm Set}\colon{\rm Set}\to{\rm Set}$ or $\operatorname{id}_{{\rm Mod}_R}\colon{\rm Mod}_R\to{\rm Mod_R}$ for a commutative unital ring $R$. These and your example are computations of the center of a category (see also here).

Answer 2

You can use the Yoneda lemma just fine here, it just takes an extra step. An endomorphism of the identity functor $\text{Grp} \to \text{Grp}$ is a compatible collection of endomorphisms of every group, whereas an endomorphism of the forgetful functor $\text{Grp} \to \text{Set}$ is a compatible collection of endofunctions of every group; in particular the former is a subset of the latter, so we can start with endomorphisms of the forgetful functor and then examine which ones also give group endomorphisms.

By the Yoneda lemma, endomorphisms of the forgetful functor correspond to endomorphisms of the representing object $\text{End}(\mathbb{Z}) \cong \mathbb{Z}$. This says that the natural endofunctions on groups all take the form

$$G \ni g \mapsto g^n \in G$$

where $n \in \mathbb{Z}$. Now the additional question we need to answer is: for what values of $n \in \mathbb{Z}$ is $g \mapsto g^n$ always a group homomorphism?

Note that if we were dealing with abelian groups then the answer is all of them, but because of the existence of nonabelian groups, here the answer is that only $n = 0, 1$ work.