Let $A$ be an abelian group. Show that $\mathrm{Hom}(\mathbb Z, A)$ is isomorphic to $A$. 
Let $A$ be an abelian group. Show that $\mathrm{Hom}(\mathbb Z, A)$ is isomorphic to $A$.

My problem is figuring out how to define Φ and using it show the homomorphism between  $\mathrm{Hom}(\mathbb Z, A)$ and $A$. 
 A: I see you are having problems in defining the homomorphism $\Phi:\mathrm{Hom}(\Bbb Z,A)\to A$, so let's deal with this in more detail. Note that $\Bbb Z$ is a group under addition $+$, and it is cyclic which means that there is an element, in this case the integer $1$, which generates each group element, i.e. every integer can be written as $1+1+\cdots+1$ or $-1-1-\cdots-1$. Now let $\phi:\Bbb Z\to A$ be a group homomorphism into an abelian group $A$. Then we have 
$$\phi(n)=\phi(\underbrace{1+\cdots+1}_{n\text{ times}})=\underbrace{\phi(1)+\cdots+\phi(1)}_{n\text{ times}}=n\cdot\phi(1)$$ as $\phi$ respects the group operation. This implies that if two homomorphisms from the integers to $A$ map the $1$ to the same image, then they are equal. In other words, if we define $$\Phi:\mathrm{Hom}(\Bbb Z,A)\to A\\\phi\mapsto\phi(1)$$ then $\Phi$ is injective. Can you show that $\Phi$ is onto?
In order to show that $\Phi$ is a group homomorphism, you just need to check that $\Phi(\phi+\psi)=(\phi+\psi)(1)=\phi(1)+\psi(1)=\Phi(\phi)+\Phi(\psi)$, but this holds by the definition of the group structure on $\mathrm{Hom}(\Bbb Z,A)$
