Why are epis stable under pullback in an abelian category ? I think the title says it all.
My question is partly motivated by the fact that this makes "element"-style reasoning with generalized elements possible; but also motivated by the result in itself. 
Here's what I attempted :
let $f: A\to B$ be an epimorphism, $g:C\to B$ any map, and assume their pullback is:
$\require{AMScd}
\begin{CD}
P @>>^i> C\\
@VV^hV @VV^gV \\
A @>>^f> B;\\
\end{CD}$
Take $\mathrm{ker}f: \mathrm{Ker}f\to A$, and an idea would be to lift this along $h$ and show that this yields someone whose cokernel is $i$: in doing so, $i$ would be an epimorphism. 
In analogy with $\mathbf{Ab}$, the kernel of $i$ would be $\mathrm{Ker}f\times \{0_C\}$ (seeing $P$ as a subgroup of $A\times C$) and so this would make sense. 
So one may consider the map $\delta: \mathrm{Ker}f \to P$ induced by $\mathrm{ker}f$ and the $0$ map to $C$ by the pullback-property ($g\circ 0 = 0 = f\circ \mathrm{ker}f$). 
I would then like to show that $i$ is the cokernel of $\delta$. But this is where I get stuck: $i\circ \delta = 0$ is clear from the definitions, but I can't see why the induced map $\mathrm{Coker}\delta \to C$ should be an isomorphism. I know it's enough to show it's a bimorphism (in an abelian category) but I can't see either part. I'm guessing that's the moment I start using the fact that $f$ is epi (I haven't until now) but I don't see how.  
 A: If the square you gave is pullback,then there is an exact sequence $0\rightarrow P\xrightarrow{(i,h)^T}C\coprod A\xrightarrow{(-g,f)}B$.since $f$ is epimorphism,we have a short exact sequence $0\rightarrow P\xrightarrow{(i,h)^T}C\coprod A\xrightarrow{(-g,f)}B\rightarrow 0$.
Hence we know the square you gave is pushout.Hence it is not difficult to show $i:P\rightarrow C$ is epimorphism.
In general,for any pullback square as you gave,the induced morphism $\overline{g}:coker(i)\rightarrow coker(f) $is injective.Hence your question has an immediate answer.
A: The answer of @Jian is clear in the first part where $P$ is constructed explicitly. However, I find the second part where $i$ is an epimorphism is not so obvious. So I will mainly focus on the proof of this part.
Since we are working in an abelian category, we can take the biproduct of $A$ and $C$, denoted by $A\oplus C$. We can define a map $\phi=(f,-g):A\oplus C \rightarrow B$ (if you prefer, you can write $\phi$ explicitly as $\phi=f\circ p_A-g\circ p_C$). Since $f$ is an epimorphism, $\phi$ is also. Then one can check that $Ker(\phi)$ is the pullback of the diagram, $i.e.$, $P=Ker(\phi)$, and we get a short exact sequence
$$0\rightarrow P\xrightarrow{j} A\oplus C\xrightarrow{\phi} B\rightarrow 0.$$
Now, the map $i:P\rightarrow C$ in the question is given by $i=p_C\circ j$. We want to prove that $i$ is an epimorphism. Let $h:C\rightarrow W$ be a map that satisfies $h\circ i=0$. Then $h\circ p_C\circ j=0$, and hence $h\circ p_C$ factors through $B$. Suppose that $h\circ p_C=r\circ \phi$ for some $r:B\rightarrow W$. Then $h\circ p_C=r\circ\phi=r\circ(f\circ p_A -g\circ p_C)$. Now, we precompose both sides with $i_A$ and get $0=r\circ f$. Since $f$ is an epimorphism, we have $r=0$ and hence $h\circ p_C=r\circ\phi=0$. Finally, since $p_C$ is an epimorphism, we get $h=0$ and we are done.
