TITLE: How to prove surjectivity part of Short Five Lemma for short exact sequences.
QUESTION [11 upvotes]: 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.

REPLY [4 votes]: Thanks to Berci and Pedro Tamaroff.
Let $b' \in B'$.  Then $\phi'(b') = c' \in C'$ and by surjectivity of $\gamma$, there's $c \in C$ with $\gamma(c) = \phi'(b')$ and there's $b \in B$ with $\phi(b) = c$.  So $\gamma\phi(b) = \phi'(b') = \phi'(\beta(b))$.  So $\beta(b) - b' \in \ker \phi' = \psi'(A') = \psi' \alpha(A)$.  So there's $a \in A$ so that $\psi'\alpha(a) = \beta(b) - b' = \beta \psi(a)$.  Then clearly $b'$ can be written as the image of the element $b - \psi(a)$ under $\beta$.  We're done.