I am trying to show that given a group $G$, there is a bijection $\mathrm{Hom}(\mathbb{Z}, G) \to G$. Here is my attempt.
Let $\Omega: \mathrm{Hom}(\mathbb{Z}, G) \to G$ by $\Omega(\phi) = \phi(1)$. We show that such a homomorphism is uniquely defined by where it sends $1$. Given $\phi, \psi \in \mathrm{Hom}(\mathbb{Z}, G)$, suppose that $\phi(1) = \psi(1)$. As $\phi$ and $\psi$ are homomorphisms, we have $\phi(0) = \psi(0) = 1_G$. Given $n > 0$, we have $$ \phi(n) = \phi(1 \cdot n) = \phi(1 + \ldots + 1) = n (\phi(1) = n \psi(1) = \psi(1 + \ldots + 1) = \psi(n). $$ Now, if $n < 0$, then $-n > 0$, so $\phi(-n) = \psi(-n)$. But homomorphisms must carry inverses to inverses, so $\phi(-n) = \phi(n)^{-1}$ and $\psi(-n) = \psi(n)^{-1}$. Therefore, $\phi(n)^{-1} = \psi(n)^{-1}$. As inverses in a group are unique, we deduce that $\phi(n) = \psi(n)$. Therefore, $\phi = \psi$. Injectivity of $\Omega$ follows immediately. Finally, we show that $\Omega$ is surjective. Given $g \in G$, define a homomorphism $\phi: \mathbb{Z} \to G$ by $\phi(1) = g$, from which the remainder of the map is uniquely determined. Then $\Omega(\phi) = \phi(1) = g$. Therefore, $\Omega$ is a bijection of sets.
How does this look?