Does this property characterize monomorphisms?

Question

Is requiring that $f:\mathrm{Hom}(A,B)$ is mono the same as requiring that the pullback $A\times_BA$ of $f$ along itself is isomorphic to $A$?

In Sheaves in Geometry and Logic, I read the following definition: $f$ being mono means that the projections out of $A\times_BA$ to $A$ (kernel pair) are both the identity. But I guess that can be weakened to saying they should be isos and the same up to iso. However, I can't quite find what "up to isomorphism" formally means for arrows (as opposed to objects, where it means they are connected by an invertible arrow). It might be that them being isos with the same codomain already means that are "the same up to iso" - could anyone clarify?

Answer 1

The following are equivalent:

• $f : A \to B$ is a monomorphism.
• The diagram below is a pullback square: $$\require{AMScd} \begin{CD} A @>{\mathrm{id}}>> A \\ @V{\mathrm{id}}VV @VV{f}V \\ A @>>{f}> B \end{CD}$$

• There exists a pullback diagram of the form below, $$\require{AMScd} \begin{CD} K @>{k_0}>> A \\ @V{k_1}VV @VV{f}V \\ A @>>{f}> B \end{CD}$$ where at least one of $k_0$ and $k_1$ is an isomorphism.

• There exists a pullback diagram of the form below: $$\require{AMScd} \begin{CD} K @>{k}>> A \\ @V{k}VV @VV{f}V \\ A @>>{f}> B \end{CD}$$

I leave the proof as an exercise for you.