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Let $\mathscr{F}$ denote the forgetful functor from the category of groups to the category of sets. Why is there more then one natural map from $\mathscr{F}$ to $\mathscr{F}$?

What are all of the natural maps from $\mathscr{F}$ to $\mathscr{F}$?

Similarly, what are the natural maps from the identity functor $\mathscr{I}:\text{Group}\rightarrow \text{Group}$ to itself?

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  • $\begingroup$ One example is given by inversion. I don't have a good argument ready for whether or not there are any others (though my guess would be that there are no others). $\endgroup$ – Tobias Kildetoft Jan 28 '20 at 17:58
  • $\begingroup$ @TobiasKildetoft Could you clarify what you mean by inversion? $\endgroup$ – 0-seigfried Jan 28 '20 at 18:00
  • $\begingroup$ I mean define the transformation $\eta$ by $\eta_G(g) = g^{-1}$ (inverse taken as in $G$). $\endgroup$ – Tobias Kildetoft Jan 28 '20 at 18:02
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The forgetful functor is representable: $\mathcal F(G)$ is naturally isomorphic to $\mathrm{Hom}(\mathbb{Z},G)$. So the question is about the natural endomorphisms of the representable functor $\mathrm{Hom}_{Gp}(\mathbb{Z},(-))$. This is immediately computed by the Yoneda lemma as $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$.

This is a very general phenomenon. Whenever any functor $U:\mathcal C\to \mathrm{Set}$ has a left adjoint $F$, one has a natural isomorphism $$U(C)\cong \mathrm{Hom}_{Set}(*,U(C))\cong \mathrm{Hom}_{\mathcal C}(F(*),C),$$ so that $U$ is representable by $F(*)$. (Here $*$ denotes a singleton set.) Then the Yoneda lemma identifies the natural endomorphisms of $U$ with the endomorphisms of $F(*)$ in $\mathcal C$. When $\mathcal C$ is some algebraic category, $F(*)$ is the free algebra on one generator, as we saw above in the case of groups.

Now for the identity functor, let $\alpha$ be a natural endomorphism of $\mathcal I$, with components $\alpha_G:G\to G$. By naturality, if $g\in G$ and $\phi_g:\mathbb Z\to G$ is the homomorphism mapping $1\mapsto g$, then we have $\alpha_G(g)=\alpha_G(\phi_g(1))=\phi_g(\alpha_{\mathbb Z}(1))=g^k$, where $k=\alpha_{\mathbb Z}(1)$. So the only possibilities are powers, as in the case of $\mathcal F$. However, the map $g\mapsto g^k$ is not a homomorphism in general, unless $k=0$ or $k=1$! This is a problem of non-abelianness, so you might continue by considering natural endomorphisms of the identity functor on abelian groups.

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  • $\begingroup$ Oh, I also found that and edited my answer... $\endgroup$ – jeanmfischer Jan 28 '20 at 20:28
  • $\begingroup$ @jeanmfischer Ah, sorry, I hadn't seen your edit before I wrote my answer. $\endgroup$ – Kevin Arlin Jan 28 '20 at 20:32
  • $\begingroup$ I finished my edit 35 seconds after your post, no worries ! $\endgroup$ – jeanmfischer Jan 28 '20 at 20:33
  • $\begingroup$ I am kind of angry with myself to not have noticed the corepresentability 1 hour ago ! :D $\endgroup$ – jeanmfischer Jan 28 '20 at 20:35
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    $\begingroup$ So for the identity on groups, the set of natural transformations is $\{0,1\}$ for abelian groups it is $\mathbb{Z}$ ! Nice ! $\endgroup$ – jeanmfischer Jan 28 '20 at 20:37
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Following this paper https://arxiv.org/pdf/1906.09006.pdf, example 3.1, it seems that $\text{Nat}(\mathscr{F},\mathscr{F})$ should be the free group on one generator, also known as $\mathbb{Z}$, the morphism of monoids $\alpha : \mathbb{Z} \to \text{Nat}(\mathscr{F},\mathscr{F})$ is given by $\alpha(n)_G:\mathscr{F}(G) \to \mathscr{F}(G), g\mapsto g^n$. Which is natural in $G$. It is clearly injective.

Actually there is a better way to see all this :

The forgetful functor $\mathscr{F}$ is corepresentable by $\mathbb{Z}$, i.e. $\mathscr{F}(G) = \text{Hom}_{\text{Grp}}(\mathbb{Z},G)$, so $$\text{Nat}(\mathscr{F},\mathscr{F}) \simeq \text{Nat}(\text{Hom}_{\text{Grp}}(\mathbb{Z},-),\text{Hom}_{\text{Grp}}(\mathbb{Z},-))\simeq \text{Hom}_{\text{Grp}}(\mathbb{Z},\mathbb{Z}) \simeq \mathbb{Z}, $$ The last isomorphism is given by the fact that any group homomorphism from $\mathbb{Z} \to \mathbb{Z}$ is given by multiplication by some $n\in \mathbb{Z}$, and the former isomorphism is given by Yoneda lemma.

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