So imagine you have a group $G$ and we consider the set of group homomorphisms from $\mathbb{Z}$ to $G$ specified by $\forall g$ $\in G$ $\exists$ $\phi(1)=g$. Each of these homomorphisms is in 1-1 relation with the elements of $G$.

Consider groups for a moment in the context of category theory, viewing the groups as one object categories and thereof group homomorphism as functors. What would a natural isomorphism mean in group theory terms for G?

My thoughts are that the natural transformations equate to maps between homomorphisms and since the homomorphisms are in 1-1 relation with the elements of G, then a natural transformation is a map between group elements, i.e another element of the group. I'm not entirely convinced by this.

Anyway continuing my reasoning a natural isomorphism is a map between the homomorphisms that has an inverse, the thing is in relation to the elements of G this is every element? Furthermore I'm looking to find an equivalence class defined by this natural isomorphism, is this just a collection of the elements that are each others inverses?

Thanks in advance for any assistance.


3 Answers 3


Generally speaking, a natural transformation $\alpha:F\Rightarrow G$ (where $F,G:\mathcal{C} \to \mathcal{D}$ are functors) is a collection of arrows $(\alpha_C:FC\to GC)$ of $\mathcal{D}$ indexed by the objects of $\mathcal{C}$ satisfying the condition that $\alpha_{C'}\circ F(c)=G(c)\circ \alpha_C$ for all $c:C\to C'$ in $\mathcal{C}$.

In your context, you only have one object, so a natural transformation between two homomorphisms $\alpha:\varphi_1 \Rightarrow \varphi_2$ is simply an arrow in the category formed by $G$, thus an element $a\in G$; the naturality condition is then simply$$a\varphi_1(n)=\varphi_2(n) a$$for all $n\in \mathbb{Z}$. Now if you take the corresponding elements of $G$, $\varphi_j(1)=g_j$ for $j=1,2$, you get that $\phi_j(n)=g_j^n$, and then the naturality can be rewritten $$g^n_2=ag_1^na^{-1}.$$

This has to hold for all $n\in \mathbb{Z}$, but it is sufficient that it holds for $n=1$.

So two functors being naturally isomorphic simply means that the corresponding elements are conjugated in $G$, and a natural isomorphism is any element that realizes this conjugation. Equivalent classes of isomorphic functors correspond to conjugacy classes in $G$.

  • $\begingroup$ Very clear answer. Do we have any analogous explanation for adjoint functors (homomorphisms)? $\endgroup$
    – Bumblebee
    Nov 30, 2020 at 6:13
  • $\begingroup$ @Bumblebee By the reasoning presented here, an adjunction is basically a pair of homomorphisms whose compositions are conjugated to the identity, or in other words, whose compositions are inner automorphism. $\endgroup$
    – Arnaud D.
    Nov 30, 2020 at 12:07

Let $G$ be a group, $a,b\in G$ be its elements, $f_a,f_b\colon\mathbb{Z}\to G$ be corresponding homomorphisms. Then for every $c\in G$, such that $b=cac^{-1}$, the assigning $*\mapsto c$ is a natural transformation $\alpha_c\colon f_a\to f_b$. Note, that every such natural transformation is a natural isomorphism, because every group is obviously a groupoid, i.e. every its morphism is an isomorphism. Hence the category $G^{\mathbb{Z}}$ of functors from $\mathbb{Z}$ to $G$ is a groupoid and its isomorphism classes are conjugacy classes of the group $G$. For example, if the group $G$ is abelian, then $$ G^{\mathbb{Z}}\cong\bigsqcup_{g\in G}G, $$ where the coproduct is taken in the category of groupoids.


I'm not sure this is the kind of thing you're looking for, but here is a categorical interpretation.

Let $Gp$ be the category of groups, and let $U: Gpe\to Set$ be the forgetful functor.

In general if $A$ is a group, then $Hom_{Gpe}(A,G)$ is not a group for all $G$, or at least not in any natural way. It's just a set. So something special happens with $A=\mathbb{Z}$.

Then your result that $Hom_{Gp}(\mathbb{Z},G) \simeq G$ can be formulated as "$\mathbb{Z}$ represents the functor $U$", ie $h_\mathbb{Z}\simeq U$ where $h_A$ is the covariant functor $G\to Set$ given by $h_A(G) = Hom_{Gp}(A,G)$. By the Yoneda lemma this characterizes $\mathbb{Z}$ up to isomorphism. Precisely, the contravariant functor $h:Gp\to C$, where $C = Fun(Gp,Set)$, given by $A\mapsto h_A$ is fully faithful.

Now a group $G$ is a group object in the category $Set$ ; this makes sense because $Set$ is a symmetric monoidal category when you endow it with the direct product (ie the cartesian product of sets).

But $C$ also has direct products : if $F,F'\in C$, then $F\times F'$ is the functor $Gp\to Set$ given by $G\to F(G)\times F'(G)$. So $C$ is also a symmetric monoidal category, and the fact that the product is computed on the level of sets means that a group object of $C$ is a functor $F: Gpe\to Set$ such that for every group $G$, $F(G)$ is a group, and that $F(G)$ is natural in $G$. In other words, a group object in $C$ is the same thing as a functor $Gp\to Gp$.

So the fact that $h_\mathbb{Z}$ is not only isomorphic to the forgetful functor $Gp\to Set$ but also to the identity functor $Gp\to Gp$ can be seen as giving a certain group object structure on $h_\mathbb{Z}$ in $C$. Now since $h:Gp\to C$ is contravariant fully faithful, and exchanges more or less by definition the product and the coproduct, a group object structure on $h_\mathbb{Z}$ is the same thing as a cogroup object structure on $\mathbb{Z}$ where $Gp$ is made into a symmetric monoidal category with respect to its coproduct (and not its product as is usual). Recall that the coproduct in $Gp$ is the so-called free product.

So we should have a comultiplication $\mu: \mathbb{Z}\to \mathbb{Z}\ast \mathbb{Z} = F_2$, a counity $\varepsilon: \mathbb{Z}\to \{1\}$ and a coinverse $i: \mathbb{Z}\to \mathbb{Z}$. It's not hard to see that $\mu(n) = (ab)^n$ where $a$ and $b$ are the generators of $F_2$, $\varepsilon$ is obviously the trivial morphism, and $i(n) = -n$.

So this gives a sort of interpretation of the fact that $Hom_{Gp}(\mathbb{Z},G)$ is naturally a group isomorphic to $G$ : the object that represents the forgetful functor $U: Gp\to Set$ is a cogroup in the category $Gp$ endowed with its coproduct, and with this additional sutrcture it represents the identity functor $Gp\to Gp$.


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