Does this property really characterize monomorphisms?

Question

In the post Does this property characterize monomorphisms?, I do not see how the third condition is equivalent to the others. Specifically, I require that $k_0$ and $k_1$ be isomorphisms in order that the third conidition is equivalent to the others. Could someone please explain this? I understand the others.

Answer 1

The third condition can only hold with $k_1=k_0$, and then their common inverse $t$ will give an isomorphism of this pullback with the one whose tip and legs are $A$ with its identities, as in the second condition.

Indeed, given a pullback square of the form in the third condition we should have a factorization $g:A\to K$ of the following square through $K$. $$\require{AMScd} \begin{CD} A @>{\mathrm{id}}>> A \\ @V{\mathrm{id}}VV @VV{f}V \\ A @>>{f}> B \end{CD}$$

Then $k_0g=\mathrm{id}_A$, so $g=t$, but $k_1g=k_t=\mathrm{id}_A$, so $k_1=k_0,$ as claimed.