Isomorphism between $\textbf{Grp}(\mathbb{Z}, G)$ and $G$

I'm learning some category after taking a break from algebra.

Could someone explain to me why, for any Group $$G$$, the set of morphisms between $$\mathbb{Z}$$ and $$G$$ is isomorphic to the group itself, that is,

$$\textbf{Grp}(\mathbb{Z}, G) \cong G$$

The group $$\mathbb{Z} = \langle 1 \mid \;\;\; \rangle$$ is the free group with one generator. That means we have a one-to-one correspondence between homomorphisms $$\mathbb{Z} \rightarrow G$$ and maps $$\lbrace 1 \rbrace \rightarrow G$$, the latter representing the choices of some element $$g \in G$$. In other words we get the map $$\varphi \colon \textbf{Grp}(\mathbb{Z}, G) \rightarrow G$$, $$f \mapsto f(1)$$ and on the other hand the map $$\psi \colon G \rightarrow \textbf{Grp}(\mathbb{Z}, G)$$, $$g \mapsto (1 \mapsto g)$$, which sends some element $$g \in G$$ to the unique homomorphism defined by the map $$1 \mapsto g$$ and explicitly given by $$a \mapsto g^a$$. By construction, we have $$\varphi = \psi^{-1}$$, which means that $$\varphi$$ (and of course also $$\psi$$) are bijections.
Recall, that the group structure on $$\textbf{Grp}(\mathbb{Z}, G)$$ is defined as $$(f \cdot g)(a) = f(a) \cdot g(a)$$, where the latter multiplication is the multiplication of $$G$$. That way we pass on the group structure of $$G$$ to $$\textbf{Grp}(\mathbb{Z}, G)$$. Now you can check that $$\varphi$$ actually not only is a bijection, but also a homomorphism of groups and thus an isomorphism.
• You should probably also say something about the group structure on $\mathbf{Grp}(\mathbb{Z}, G)$. Sep 17 '19 at 19:33
• can you clarify definition of $\psi$? What do you mean by $g \mapsto (1 \mapsto g)$ ? Sep 17 '19 at 20:12