Understand one line of proof in Homological Algebra In Saunders Maclane's book "Homology". There is a line which states that if $(\alpha, \beta, \gamma)$ is a map from one short exact sequence to another. Then $\alpha, \beta$ monic implies $\gamma$ monic by the composition $(\alpha, \beta, \gamma) \circ (\alpha', \beta', \gamma') = (\alpha \alpha', \beta \beta', \gamma \gamma')$, where the primes are also map a short exact sequence to another.
I do not see how $\gamma$ is monic by this argument.
Edit: Maclane says $\alpha, \beta$ monic implies $(\alpha, \beta, \gamma)$ is monic as a map of exact sequences, not $\gamma$ monic.
 A: Could you give the precise page in Mac Lane's book? There is probably a typo. You are claiming that in a commutative diagram with exact rows
$$\require{AMScd}\begin{CD}
0 @>>> A' @>>> A @>>> A'' @>>> 0 \\
@. @V \alpha VV @V \beta VV @V \gamma VV @. \\
0 @>>> B' @>>> B @>>> B'' @>>> 0 \\
\end{CD}$$
if $\alpha$ and $\beta$ are mono, then $\gamma$ is mono. This is not true. Here's a counterexample:
$$\require{AMScd}\begin{CD}
0 @>>> \mathbb{Z} @>\times 4>> \mathbb{Z} @>>> \mathbb{Z}/4\mathbb{Z} @>>> 0 \\
@. @V \times 2 VV @V id VV @VVV @. \\
0 @>>> \mathbb{Z} @> \times 2 >> \mathbb{Z} @>>> \mathbb{Z}/2\mathbb{Z} @>>> 0 \\
\end{CD}$$
Recall the snake lemma which says that there is an exact sequence
$$0 \to \ker \alpha \to \ker \beta \to \ker \gamma \to \operatorname{coker} \alpha \to \operatorname{coker} \beta \to \operatorname{coker} \gamma \to 0$$
(see section II.5 in Mac Lane's book).
So what is true is that if $\alpha$ and $\gamma$ are mono (resp. epi, iso), then $\beta$ is mono (resp. epi, iso).

Now that you edited the question, the statement makes sense: if
$$\require{AMScd}\begin{CD}
0 @>>> A' @>>> A @>>> A'' @>>> 0 \\
@. @V \alpha VV @V \beta VV @V \gamma VV @. \\
0 @>>> B' @>>> B @>>> B'' @>>> 0 \\
\end{CD}$$
is a morphism of short exact sequences with $\alpha$ and $\beta$ being mono, then it is a mono.
Indeed, we should show that in this case
$$(\alpha,\beta,\gamma)\circ (\alpha',\beta',\gamma') = (\alpha\circ\alpha', \beta\circ\beta', \gamma\circ\gamma') = (0,0,0)$$
implies $(\alpha',\beta',\gamma') = (0,0,0)$.
So we have a commutative diagram
$$\require{AMScd}\begin{CD}
0 @>>> C' @>>> C @>>> C'' @>>> 0 \\
@. @V \alpha' VV @V \beta' VV @V \gamma' VV @. \\
0 @>>> A' @>>> A @>>> A'' @>>> 0 \\
@. @V \alpha VV @V \beta VV @V \gamma VV @. \\
0 @>>> B' @>>> B @>>> B'' @>>> 0 \\
\end{CD}$$
with $\alpha\circ \alpha' = 0$, $\beta\circ \beta' = 0$, $\gamma\circ \gamma' = 0$. As $\alpha$ and $\beta$ are mono, this implies $\alpha' = \beta' = 0$. But then $\gamma' = 0$, being the morphism on cokernels induced by $\alpha'$ and $\beta'$.
Similarly, if $\beta'$ and $\gamma'$ are epi, then $(\alpha',\beta',\gamma')$ is epi in the category of short exact sequences.
This gives an example of a morphism of short exact sequences which is both mono and epi, but not iso:
$$\require{AMScd}\begin{CD}
0 @>>> 0 @>>> A @> id >> A @>>> 0 \\
@. @V 0 VV @V id VV @V 0 VV @. \\
0 @>>> A @> id >> A @>>> 0 @>>> 0 \\
\end{CD}$$
Here the first two vertical arrows are mono, and the last two arrows are epi. But $(\alpha,\beta,\gamma)$ is an iso in the category of short exact sequences if and only if $\alpha$, $\beta$, $\gamma$ are all iso.
