Example 3.5 in Allufi: Chapter 0 (Slice categories) 

Now on the final line is what is confusing me. What I understand by what the author is saying is that the morphism $\sigma \in \text{Hom}_{C_A}(f_1, f_2)$ is exactly the morphism $\sigma \in \text{Hom}_C(Z_1, Z_2)$ such that $f_1 = f_2 \circ \sigma$. 
It is clear to me why $\sigma$ exists (because $C$ is a category and every morphism in $C$ has a source and a target object). However it is not clear to my why $f_2 \circ \sigma = f_1$. In the definition of a category there exists a mapping called composition, written by $(f, g) \mapsto g \circ f$, so $f_2 \circ \sigma$ certainly exists, but nothing in the definition of the category $C$ states that we need to have $f_1 = f_2 \circ \sigma$. 
Also note that in the definition of a category, the mapping called composition is not surjective. So since $f_1 \in \text{Hom}_C(Z_1, A)$, there need not exist a $f_2 \circ \sigma$ with $f_2 \in \text{Hom}_C(Z_2, A)$ and $\sigma \in \text{Hom}_C(Z_1, Z_2)$ such that $(\sigma, f_2) \mapsto f_2 \circ \sigma = f_1$.
So what exactly does the author mean?
 A: It is true that an arrow $\sigma : Z_1 \to Z_2$ need not satisfy $f_1 = f_2 \circ \sigma$. If it doesn't, then it is not a member of $\hom_{C_A}(f_1, f_2)$. If it does, then it is a member of $\hom_{C_A}(f_1, f_2)$.
It can indeed be the case that there aren't any $\sigma : Z_1 \to Z_2$ at all that satisfy $f_1 = f_2 \circ \sigma$. In this case, $\hom_{C_A}(f_1, f_2) = \varnothing$.
A: I will denote the slice category $\mathbf{C}$ over $A$ as $\mathbf{C}\downarrow A$.
You are looking at it from the wrong angle, I suppose. He is saying that if there exists a morphism $\sigma\in \mathrm{Hom}_\mathbf{C}(Z_1, Z_2)$ that makes the triangle commute ($f_1 = f_2\circ\sigma$), then we define a canonical morphism $\hat\sigma\in\mathrm{Hom}_{\mathbf{C}\downarrow A}(f_1, f_2)$, for $f_i:Z_i\rightarrow A$. In particular that homset in $\mathbf{C}\downarrow A$ could possibly be empty.
Phrased differently, he only defines morphism in the slice category via morphisms that make triangles commute - others are not taken into consideration, you don't need to prove anything. 
Well, of course you still have to check that $\hat{\sigma}$ satisfies the definition of a morphism, but that is quite easy, and you should definitely do that as an exercise ;)
