Are commutative squares that induce iso on cokernels cocartesian?

Question

Does the converse of this question hold true? Explicitly, is a commutative square $$\begin{matrix} A & \stackrel a \to & X\\\llap{\scriptstyle a'}\downarrow&&\downarrow\rlap{\scriptstyle b'}\\Y & \stackrel b \to & B \end{matrix}$$ cocartesian if the induced maps $\operatorname{coker}a \to \operatorname{coker}b$ and $\operatorname{coker}a' \to\operatorname{coker}b'$ are isomorphisms?

Actually, I see no reason why this should be true, but maybe someone has a nice counter example and/or can elaborate on the problem.

Answer 1

Here is an example: in the diagram below, the top and left arrows send $1 \mapsto 1$; then all cokernels are $0$, but the pushout $\mathbb{Z}/2 \sqcup_{\mathbb{Z}} \mathbb{Z}/2$ is $\mathbb{Z}/2$. $\require{AMScd}$ \begin{CD} \mathbb{Z} @>>> \mathbb{Z}/2 \\ @VVV @VVV\\ \mathbb{Z}/2 @>>> 0 \end{CD}