Quotients preserve monomorphisms in abelian categories We work in an abelian category. I use $f_k$ and $f^k$ to denote the morphisms given by the kernel and cokernel of the map $f$.
Here's a diagram
Suppose we have monos $f:A\rightarrow B$, $g:A\rightarrow C$, and $h:B\rightarrow C$. Then $g^khf=g^kg=0$, so $g^kh$ factors uniquely through $f^k$, say via some $h'$. That is, $h'f^k=g^kh$. Must $h'$ be mono?
I want to say yes after mulling it over in the category of modules over a ring, in light of Mitchell's embedding theorem. Is there also an abstract nonsense proof? I'm not really sure how to translate my proof using elements into one about morphisms.
Probably there's also a proof involving commuiting limits and colimits of a certain kind, but I don't know enough about this to come up with the proof myself (but if you do, feel free to share).
 A: Let $Y$ be any object, with a morphism $u\colon Y\to B/A$ such that $h'u=0$.  We must show that $u=0$, to deduce that $h'$ is monic.

In the special case where the abelian category has enough projectives you can proceed as follows:
Let $P$ be a projective object with epic morphism $p\colon P\to Y$.  By the projective property of $P$ we have a morphism $v\colon P \to B$ such that $f^kv=up$.

In the general case, let $P$ and the maps $p\colon P\to Y$, $v\colon P \to B$ be the pullback of $B$ and $Y$ over $B/A$.  Such a pullback exists and $p$ will be epic as it is the pullback of $f^k$ (see here).  By commutativity of the pullback square we have $f^kv=up$.

Either way, to deduce that $u=0$ it will suffice to show that $up=0$ (as $p$ is epic).  Our goal now is to show that $v$ factors through $f$ (so $v=fm$), as then we have $up=f^kv=f^kfm=0$.

We have $g^khv=h'f^kv=h'up=0$.
As $g$ is monic, we have $g=(g^k)_k$, so $g^khv=0$ implies $hv=gm$ for some morphism $m\colon P\to A$.
Then $hv=gm=hfm$.  As $h$ is monic we may conclude that $v=fm$ as was our goal.
