Is this apushout diagram [closed]

Question

Let $A, B, C, E$ and $F$ be some objects in an abeleian category $\mathcal{C}$. Let we have a commutative diagram

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

where the first downarrow ids an isomorphism and the second is a monomorphism. Then 1- Is it true to say the the third downarrow (i.e. $C\to F$) is a monomorphism? 2- Is it true to say that the right square is a pushout diagram?

Answer 1

I'm assuming you mean the rows to be short exact sequences, as well. Then

1- Yes, by the Snake lemma.

2- Yes. The square is a pushout if and only if the square $$\require{AMScd} \begin{CD} B @>(q,-i)>> C \oplus E \\ @VVV @VVj+rV \\ 0 @>>> F \end{CD} $$ is a pushout, where $i\!: B \hookrightarrow E$ and $j\!: C \hookrightarrow F$ are the maps from the diagram. This in turn is equivalent to the sequence $B \xrightarrow{(q,-i)} C \oplus E \xrightarrow{j+r} F \to 0$ being exact, which is routine to check.

In this situation, the squares are actually also pullbacks, and the sequence above is exact on the left, as well (since $i$ is a monomorphism).