Consider the following diagram which lives in the category of $R$-modules.


$$
\begin{array}{ccccccccc}
0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0 \newline
& & \downarrow & & \downarrow & & \downarrow & & \newline
0 & \xrightarrow[j]{} & A & \xrightarrow[g]{} & E & \xrightarrow[r]{} & F & \xrightarrow[e]{} & 0
\end{array}
$$


Let the first down arrow be an equivalence and the other down arrows be both epimorphisms.


Can we prove that the right square is a push out diagram?
If not what can be said about the kernels of these two maps? Are they isomorphic?


What can be said about the diagram which is mirror revers of this diagram?



Is this square a push-out square?
The right square is a pushout square as soon as the first downward arrow is an epimorphism, and we do not even need $f$ and $g$ to be monomorphisms. This is true in any abelian category. 


Since diagrams are not easy to draw, let me label the downward arrows as $a : A \to A'$ (which we assume is an epimorphism but not necessarily an isomorphism), $b : B \to E$ and $c : C \to F$. Suppose $h\_1 : E \to G$ and $h\_2 : C \to G$ are morphisms such that $h\_1 \circ b = h\_2 \circ q$. Then, $h\_1 \circ b \circ f = h\_2 \circ q \circ f = 0$, so $h\_1 \circ g \circ a = 0$ (because $g \circ a = b \circ f$). Since $a$ is an epimorphism, we must have $h\_1 \circ b = 0$, thus $h\_1 = h \circ r$ for a unique $h$, by the universal property of cokernels. Moreover, $h \circ c \circ q = h \circ r \circ b = h\_1 \circ b = h\_2 \circ q$, so $h \circ c = h\_2$, because $q$ is an epimorphism. So we indeed have a pushout square.


