Any straightforward proof of "in an abelian category, a pullback yields a monomorphism at cokernel level"? Here is the question I encountered:

$$\require{AMScd}
\begin{CD}
s @>{f^\prime}>> a @>{\varphi^\prime}>> \bar a\\
@V{g^\prime}VV @V{g}VV @V{\bar g}VV\\
b @>{f}>> c @>{\varphi}>> \bar c
\end{CD}$$
Given a commutative diagram as above in an abelian category such that $(f^\prime,g^\prime)$ is a pullback of $(f,g)$ and  that $\varphi$, $\varphi^\prime$ are cokernel of $f$, $f^\prime$ respectively. Then  $\bar g$ is monic.

I proved it using diagram chasing but I wonder if there is any simpler and more straightforward proof. I view this as a small generalization of the fact that a pullback of an epimorphism is epic in an abelian category, I think it's highly possible to find one, but I haven't succeeded yet. Does anyone have any idea?
 A: Recall the fact that pullbacks preserves kernels,
then we can deduce from this fact that the map on cokernels is a monomorphism
by using a spectral sequence argument.

First, we may regard the following pullback square as a double complex :
$$\require{AMScd}
\begin{CD}
b @>{f^\prime}>> c\\
@A{g^\prime}AA @AA{g}A\\
s @>{f}>> a
\end{CD}$$
Computing rightwards, then the first two pages are given by
$$\newcommand{\on}{\operatorname}
_{>}E^1_{\bullet\bullet}:\quad\begin{CD}\on{Cok}(g')@>>>\on{Cok}(g)\\\\\on{Ker}(g')@>>>\on{Ker}(g)\end{CD}\quad\quad\quad{_{>}E^2_{\bullet\bullet}}:\quad\begin{CD}?&&&&?\\\\0&&&&0\end{CD}$$
(the arrows in $_>E^2_{\bullet\bullet}$ are omitted)
From this, we see that $_>E^{\infty}_{\bullet\bullet}={_>E^{2}_{\bullet\bullet}}$, and that there are two $0$'s at the bottom of $_>E^{\infty}_{\bullet\bullet}$.
Similary, if we compute upwards, the first two pages ought to be :
$$_{\wedge}E^1_{\bullet\bullet}:\quad\begin{CD}\on{Ker}(f') &&&& \on{Cok}(f)\\@AAA && @AAA\\\on{Ker}(f')&&&& \on{Cok}(f)\end{CD}\quad\quad\quad{_{\wedge}E^2_{\bullet\bullet}}:\quad\begin{CD}0&&&&?\\\\0&&&&\star\end{CD}$$
and similarly that $_{\wedge}E^{\infty}_{\bullet\bullet}={_{\wedge}E^{2}_{\bullet\bullet}}$.
If we can show that $\star=0$, then we are done.
Since the original double complex has finitely many nonzero terms, we are done by noticing
$$_{>}E^{2}_{\bullet\bullet}={_{>}E^{\infty}_{\bullet\bullet}}={_{\wedge}E^{\infty}_{\bullet\bullet}}={_{\wedge}E^{2}_{\bullet\bullet}}$$

Other examples on proving some diagrammatic results in abelian categories by spectral sequences
can be found in This note on Spectral Sequence by Vakil.
