STATEMENT: This is taken from Robert Ash's,Basic Abstract Algebra. Let us call $0$ the zero object in an arbitrary category. And let us denote $0_{AB}$ the zero morphism from an object $A$ in the category to an object $B$ in the same category. Why is it that given any arbitrary morphism, $h$, in this category that $h0_{AB}=0_{AB}h=0_{AB}$.
EXACT QUOTE FROM BOOK: "Note that since a zero morphism goes through a zero object, it follows that for an arbitrary morphism $h$, we have $h0=0h=0$."
QUESTION: I don't get how the last relationship with the morphism $h$ is derived. I am assuming that he assumes $A=B$ otherwise the thing he states can't necessarily hold.
A zero object $0$ in a category is an object that is both initial and terminal. Initial means there exists a unique morphism from this object to any other given object $A$, call it $0_{0A}: 0 \rightarrow A$. Terminal means there exists a unique morphism from any given object $A$ to this object, call it $0_{A0}: A \rightarrow 0$.
If our category contains a zero object $0$, given any two other objects $A$ and $B$ we can define the zero morphism $0_{AB}$ to be $A \rightarrow 0 \rightarrow B$, where the arrows are the unique morphisms. That is, $0_{AB} = 0_{0B} \circ 0_{A0}$.
Now what if you have an arbitrary morphism $h: B \rightarrow C$? Claim: $h \circ 0_{AB} = 0_{AC}$. Proof: $h \circ 0_{AB} = h \circ 0_{0B} \circ 0_{A0}$. The morphism $h \circ 0_{0B}: 0 \rightarrow C$ must equal $0_{0C}$ by uniqueness (i.e. because $0$ is an initial object). Then $h \circ 0_{AB} = 0_{0C} \circ 0_{A0} = 0_{AC}$.
Now suppose $h: C \rightarrow A$. Claim: $0_{AB} \circ h = 0_{CA}$. Prove this one yourself (the proof is very similar to the above paragraph).