A commutative square is called bicartesian when it is both pull-back and push-out. In an abelian category, consider two pull-back squares $(X)$ and $(Y)$: $$ \begin{array}{ccccc} A & \longrightarrow & B & \longrightarrow & C \\ \downarrow & (X) & \downarrow & (Y) & \downarrow \\ D & \longrightarrow & E & \longrightarrow & F \end{array} $$
Prove that:
The square $(X+Y)$ is a bicartesian square $\Longleftrightarrow$ The squares $(X)$ and $(Y)$ are bicartesian squares.
I have been able to prove, R.H.S $\Longrightarrow$ L.H.S but I am not being able to prove the other way. Please help.
Let's start by labeling the arrows:$\require{AMScd}$
$$ \begin{CD} A @>f>> B @>g>> C\\ @VVaV @VVbV @VVcV\\ D @>>h> E @>>k> F \end{CD} $$
The square $(X+Y)$ is exact if and only if the "diagonal sequence" $0 \to A \xrightarrow{(gf, a)^t} C \oplus D \xrightarrow{(-c,kh)} F \to 0$ is exact.
The squares $(X)$ and $(Y)$ are pull-backs if and only if the "diagonal sequences" $0 \to A \xrightarrow{(f,a)^t} B \oplus D \xrightarrow{(b,-h)} E$ and $0 \to B \xrightarrow{(g,b)^t} C \oplus E \xrightarrow{(c,-k)} F$ are (left) exact.
So, assuming that $(X+Y)$ is exact and $(X)$ and $(Y)$ are pull-backs, we can conclude that $(Y)$ is exact since $(c,-k) \circ (1\oplus h) = (c,-kh)$ is epic, so $(c,-k)$ is epic and hence $(Y)$ is exact.
Now we know that $(X+Y)$ and $(Y)$ are both exact and it follows that so is $(X)$ by the fact mentioned by tetrapharmakon.