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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.

Thanks in advance

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  • $\begingroup$ If two objects represent the same functor, they must be isomorphic. $\endgroup$ – Angina Seng Mar 27 '19 at 20:42
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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.

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  • $\begingroup$ 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 ?? $\endgroup$ – Pedro Mar 27 '19 at 20:58
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    $\begingroup$ @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 $n\cdot f(1) = f(n\cdot 1) = f(0)$. $\endgroup$ – lisyarus Mar 27 '19 at 22:54
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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 $1$.

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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.

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