Zero arrow is independent of zero object

Question

In a category $\mathcal{C}$ a zero object $0\in\mathcal{C}$ is an object such that there is for every $A\in\mathcal{C}$ precisely one element in $\mathcal{C}(A,0)$ and $\mathcal{C}(0,A)$, called the zero arrows. Let us denote them by $0_{A,0}$ and $0_{0,A}$.

By composition we get a unique zero arrow $0_{A,B}$ for $A,B\in\mathcal{C}$ by $0_{0,B}\circ 0_{A,0}$.

Besides one can show that any two zero objects are isomorphic by an unique isomorphism.

But I was told that the zero arrow $0_{A,B}$ is independent of the choice of the zero object, i.e. if $Z$ is another zero object, than $0_{A,B}=Z_{A,B}$.

Can somebody explain, why?

Answer 1

If $Z$ is another zero object and $\varphi$ is the unique isomorphism between $0$ and $Z$, then the following diagram will commute:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & A & \ra{0_{A,0}} & 0 & \ra{0_{0,B}} & B & \\ & \da{\text{id}} & & \da{\varphi} & & \ua{\text{id}} \\ & A & \ra{Z_{A,0}} & Z & \ra{Z_{0,B}} & B & \\ \end{array} $$ Edit: by commutativity of the smaller squares, we can calculate

$$\begin{align*} Z_{A,B} &= Z_{0,B}\circ Z_{A,0} \\ &= \text{id}\circ Z_{0,B}\circ Z_{A,0}\circ\text{id}\\ &= 0_{0,B}\circ\varphi^{-1}\circ Z_{A,0}\circ\text{id}\\ &= 0_{0,B}\circ\varphi^{-1}\circ\varphi\circ 0_{A,0}\\ &= 0_{0,B}\circ 0_{A,0}\\ &= 0_{A,B} \end{align*}$$

which is the result you wanted. Morally, what happened is that seeing that the smaller squares commuted was easy by the universal property of the zero object. This will always imply that putting them together to get a "larger square" preserves commutativity. So writing the commutative diagram "replaces" the work done by writing out the long list of function equalities like I just did.