Suppose we have a homomorphism $\alpha, \beta, \gamma$ of short exact sequences:
$$
\begin{matrix}
0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & 0 \\
\ & \ & \downarrow^{\alpha} & \ & \downarrow^{\beta} \ & \ & \downarrow^{\gamma} \\
0 & \to & A' & \xrightarrow{\psi'} & B' & \xrightarrow{\phi'} & C' & \to & 0
\end{matrix}
$$
If both $\alpha, \gamma$ are surjective then so is $\beta$. This can be proved using the properties of the diagram somehow.
I've tried several things.
Best Answer
It should be straightforward.
We want to prove $\beta$ is surjective, so start out from an arbitrary element $b'\in B'$. We can do one thing: consider $c':=\phi'(b')\in C'$.
Since $\gamma$ is surjective, we get $c$ with $\gamma(c)=c'$.
That the pair $\phi,0$ of maps is exact means nothing else but that $\phi$ is surjective. It yields an element $b\in B$, such that $\gamma\phi(b)=c'=\phi'(b')$.
Now we might not get $\beta(b)=b'$ with this element $b$, nevertheless, we have that $\beta(b)$ and $b'$ has the same image under $\phi'$, so by exactness, this gives $a'\in A'$ such that $\psi'(a')=b' - \beta(b)$.
Can you take it on from here?