Working my way through Aluffi's "Algebra: Chapter $0$", he mentions that the set of homomorphisms of abelian groups form an abelian group themselves. He doesn't go much into details at that point, but I started thinking about homomorphisms between such groups, specifically ones with common "target": $\text{Hom}_{\text{Ab}}(A,C) \rightarrow \text{Hom}_{\text{Ab}}(B,C)$.
Let $\alpha \in \text{Hom}_{\text{Ab}}(A,C)$ and $\beta \in \text{Hom}_{\text{Ab}}(B,C)$. Initially I thought homomorphisms between the $\text{Hom}$s would correspond to homomorphisms $\varphi : A \rightarrow B$ such that:
commutes, i.e. $\alpha = \beta \circ \varphi$. (Getting the inspiration from how he described slice categories)
I say "initially" since I also started thinking about the kernel of non-trivial $\varphi$ and it seems to me that it must always be trivial under this definition? (Too good to be true?)
My thinking was this: Letting $\beta$ be the homomorphism that sends everything to $0 \in C$; if $\varphi$ itself is not trivial, the only $\alpha$ that would make the diagram commute would be the one that sends everything to $0 \in C$ too.
What I'm not sure is if my misunderstanding comes from how I'm thinking about the homomorphims in terms of such diagrams or, if the previous view is correct, I'm just missing how a non-trivial $\alpha$ could commute with a trivial $\beta$. (Or it could be something completely different, of course.)
Thanks in advance!