If for monos $u\leq v$ and $v\leq u$ then their domains are isomorphic

Question

I'm unable to prove that if a mono $u:B\to A$ is less then mono $v:C\to A$ by $f:B\to C$ with $v\circ f=u$ and also $v\leq u$ by $u\circ g=v$ then $B\cong C$ by using some morphisms as above, but I believe that this holds. Is there a hint how this $\cong$ may look like together with a hint that it is $\cong$ ? This is certainly true in the category of Sets and functions but I suspect that it holds in general. Also, I think that this is further more equivalent to $u=v\circ \theta$ for some $\theta$ being an isomorphism. How can I see this equivalence ?

Answer 1

Let $u \colon B \to A$ and $v \colon C \to A$ monomorphisms in a category. Then the following are equivalent:

1. There are morphisms $f \colon B \to C$ and $g \colon C \to B$ such that $u=vf$ and $v=ug$.
2. There is an isomorphism $\theta \colon B \to C$ such that $u=v\theta$.

Proof: If we assume (1), it suffices to prove that $f$ is an isomorphism (so, $\theta=f$ works for (2)). Indeed, since $$ u\text{id}_B=u=vf=(ug)f=u(gf) $$ and $u$ is mono, then $gf=\text{id}_B$. Also, since $$ v\text{id}_C=v=ug=(vf)g=v(fg) $$ and $v$ is mono, then $fg=\text{id}_C$. Hence, $f$ is an isomorphism.

On the other hand, if we assume (2), then the morphisms $f=\theta$ and $g=\theta^{-1}$ works for (1).