# Why does $\mathbb{Z}$ represent the forgetful functor $U:\mathbf{Grp}\to\mathbf{Set}$

This is from Emily Riehl's Category theory in context

The forgetful functor $$U:\mathbf{Grp}\to\mathbf{Set}$$ is represented by the group $$\mathbb{Z}$$ thanks to the natural isomorphism $$\alpha:\mathbf{Grp}(\mathbb{Z},-)\cong U$$ whose components are the isomorphisms \begin{align} \alpha_G:\mathbf{Grp}(\mathbb{Z},G) & \longrightarrow UG\\ \big[f:\mathbb{Z}\to G\big] & \longmapsto f(1) \end{align} My impression is that if some group $$A$$ were to represent the functor $$U$$ it should have some distinguished element we could attach all the information to. If the group $$A$$ doesn't have such element then the only "reasonable" map $$\mathbf{Grp}(A,G)\to UG$$ would be $$\big[f:A\to G\big]\mapsto f(id_A)$$ and that does not define an isomorphism. The only groups having a "distinguished element" that come to mind are the cyclic groups and their generators.

But then, why $$\mathbb{Z}$$? Doesn't every (non trivial) cyclic group work? Given the cyclic group $$\mathbb{Z}_n$$, the components \begin{align} \alpha_A:\mathbf{Grp}(\mathbb{Z_n},G) & \longrightarrow UG\\ \big[f:\mathbb{Z}_n\to G\big] & \longmapsto f(1) \end{align} are isomorphisms: if $$f(1)=\alpha_A(f)=\alpha_A(g)=g(1)$$ then, for every $$m\in\mathbb{Z}_n$$ $$f(m)=f(1)+\cdots+f(1)=g(1)+\cdots+g(1)=g(m)$$ so that $$g\equiv f$$ and for every $$g\in UG$$ there exists a unique $$f_g:\mathbb{Z}_n\to G$$ such that $$f_g(1)=g$$.

Where am I wrong? Could it be that $$\mathbb{Z}=\langle1\rangle$$ is the only cyclic group with only one generator while $$\mathbb{Z}_n=\langle1\rangle=\langle n-1\rangle$$ for every $$n$$? If so, why? It doesn't seem very important to the problem.

Take $$G=\Bbb Z$$. Then the only element of $${\bf Grp}(\Bbb{Z}_m,\Bbb Z)$$ is the zero map. But $$U\Bbb Z$$ has rather more than one element.
• so the problem is that defining only the image of $1\in\mathbb{Z}_n$ does not necessarily define a homomorphism $\mathbb{Z}_n\to$something ?? – Pedro Mar 27 '19 at 20:58
• @Pedro Exactly. For a homomorphism $f: \mathbb Z_n \rightarrow G$ the image of $1 \in \mathbb Z_n$ should have order dividing $n$, for $n\cdot f(1) = f(n\cdot 1) = f(0)$. – lisyarus Mar 27 '19 at 22:54
Let $$S_1 = \{ 1 \}$$. Then you have a natural bijection between $$\mathbf{Set}(S_1,UG)$$ and $$UG$$. Moreover, for any set $$S$$, you have a natural bijection between $$\mathbf{Set}(S,UG)$$ and $$\mathbf{Grp}(F(S),G)$$, where $$F(S)$$ is the free group with basis $$S$$. This is the universal property of the free group generated by $$S$$. But $$F(S_1)$$ is nothing else than the infinite cyclic group with one generator $$1$$.
When you have an adjunction $$\mathsf{F}: \mathsf{Set} \leftrightarrows \mathsf{K}: \mathsf{U},$$ where $$\mathsf{U}$$ is a forgetful like functor, $$\mathsf{U}$$ is always representable. This is due to a very specific entanglement that is characteristic of the category of sets (and in general will kinda apply for $$\mathcal{V}$$ in $$\mathcal{V}$$-$$\mathsf{Cat}$$ when $$\mathcal{V}$$ is monoidal closed). In fact $$\mathsf{U}(\_) \cong \mathsf{Set}(1, \mathsf{U}(\_)) = \mathsf{K}(\mathsf{F}1, (\_)),$$ and thus the forgetful functor is representable and is represented by the free algebra over the one element set. This is evident when $$\mathsf{K}$$ is $$\mathsf{R}$$-$$\mathsf{Mod}$$, groups, monoids and algebraic structures in general.