# Self-Natural Maps for Forgetful Functor

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?

• 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). – Tobias Kildetoft Jan 28 '20 at 17:58
• @TobiasKildetoft Could you clarify what you mean by inversion? – 0-seigfried Jan 28 '20 at 18:00
• I mean define the transformation $\eta$ by $\eta_G(g) = g^{-1}$ (inverse taken as in $G$). – Tobias Kildetoft Jan 28 '20 at 18:02

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.

• Oh, I also found that and edited my answer... – jeanmfischer Jan 28 '20 at 20:28
• @jeanmfischer Ah, sorry, I hadn't seen your edit before I wrote my answer. – Kevin Arlin Jan 28 '20 at 20:32
• I finished my edit 35 seconds after your post, no worries ! – jeanmfischer Jan 28 '20 at 20:33
• I am kind of angry with myself to not have noticed the corepresentability 1 hour ago ! :D – jeanmfischer Jan 28 '20 at 20:35
• So for the identity on groups, the set of natural transformations is $\{0,1\}$ for abelian groups it is $\mathbb{Z}$ ! Nice ! – jeanmfischer Jan 28 '20 at 20:37

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.