The following is a commutative diagram in an abelian category. Assume that the rows are exact and that $h,k$ are epic.
$\require{AMScd}$
$$\begin{CD}
0@>>>a@>{f}>> b @>{g}>> c@>>> 0\\
@VVV @VVjV @VVhV @VVkV @VVV\\
0@>>>a'@>{f'}>> b' @>{g'}>> c'@>>> 0
\end{CD}$$
How do I prove that $j$ is epic if the square $g,h,k,g'$ is cocartesian? I proved the converse already, but I'm stuck in this direction. Thanks in advance.
EDIT. I post this solution for completeness, since it seems different from the one given in the answer. I didn't use that $h,k$ are epic, and this hypothesis isn't useful in proving the converse of sentence in italics either, so I don't know why it was in text of the exercise.
Let $z:a'\to d$ be an arrow such that $zj=0$, and construct the pushout: $$\begin{CD}
a'@>{f'}>> b'\\
@VVzV @VVp_1V \\
d@>{p_2}>> p
\end{CD}$$ since $p_1hf=0$ and $g=\operatorname{coker}f$, there is $u:c\to p$ s.t. $ug=p_1h$. Being $g,h,k,g'$ a pushout, there is $v:c'\to p$ s.t. $vg'=p_1$, so that $p_2z=p_1f'=vg'f'=0$. Since $p_2$ is monic (it is the pushout of $f'$, that is monic), $z=0$.
Best Answer
Consider the following commutative diagram: \begin{CD} a@>>>0\\ @VfVV @VVV\\ b @>{g}>> c\\ @VVhV @VVkV\\ b' @>{g'}>> c' \end{CD} By exactness of top row, $g$ is the cokernel of $f$, hence the top square is a pushout. By pushout pasting lemma, the bottom square is a pushout if and only if the outer rectangle is a pushout as well. The outer rectangle is a pushout if and only if $g'$ is the cokernel of $fh$. By exactness of the top row, $f'$ is the kernel of $g'$. Consequently, $g'$ is the cokernel of $fh$ if and only if $f'$ is the image of $fh$. Finally, $j$ is epic if and only if $f'$ is the image of $jf'$, hence that of $fh$.