The following is taken from $\textit{Module Theory An Approach to Linear Algebra}$ By: T.S.Blyth
$\color{Green}{Background:}$
$\textbf{Theorem 3.9:}$ $\textbf{[The four lemma]}$ Suppose that the diagram of $R-$modules and $R-$morphisms:
$$\begin{array}{ccccccccc} A & \xrightarrow{f} & B & \xrightarrow{g} & C \xrightarrow{h} & D \\ \big\downarrow\alpha & & \big\downarrow\beta & & \big\downarrow \gamma & \big\downarrow \delta & \\ A' & \xrightarrow{f'}& B' & \xrightarrow{g'} & C' \xrightarrow{h'} & D'\end{array}$$
is commutative and has exact rows. Then we have
$(1)$ if $\alpha, \gamma$ are epimorphisms and $\delta$ is a monomorphism then $\beta$ is an epimorphism
$(2)$ If $\alpha$ is an epimorphism and $\beta, \delta$ are monomorphisms then $\gamma$ is a monomorphism
$\textit{Proof.}$ $(1)$ Let $b'\in B'.$ Since $\gamma$ is surjective there exists $c\in C$ such that $g'(b')=\gamma(c).$ By the commutativeity of the right-hand square we then have $\delta[h(c)]=h'[\gamma(c)]=h'[g'(b')]=0,$ since $h'\circ g'=0.$ Thus:
$\color{Red}{Questions:}$
I have a quick question about the partial proof of the Theorem above. The line
"$\delta[h(c)]=h'[\gamma(c)]=h'[g'(b')]=0$ since $h'\circ g'=0$".
What is the reason that we can conclude
"$\delta[h(c)]=h'[\gamma(c)]=h'[g'(b')]=0,$"?
Is it because
$(1)$ Commutativity of the square
$(2)$ Commutativity of the square and that $\delta$ is a epimorphism, and hence the string of equalities allow for the map $D\xrightarrow{\delta}D'$ to be extended to the zero morphism $D'\to 0$ in order to make it exact, so that $\delta\circ h=0$ and in turn $h'\circ g'=0?$
Thank you in advance
It is too much back and forth in the comments. The answer to the original question is commutativity of the right square and exactness at $C'$, but the question now seems to be what exactness means.
Given a sequence of abelian groups or $R$-modules (this can be more general):
$A\xrightarrow{f}B\xrightarrow{g}C\space$ (this can be part of a longer sequence including an infinite one)
the sequence is exact at $B$ if $Im\space f = Ker\space g$.
$Im\space f \subseteq Ker\space g$ is equivalent to $g\circ f =0$.
$Ker\space g \subseteq Im \space f$ means that any element sent to $0$ by $g$ comes from some element in $A$ via $f$.
$A$ or $C$ being $0$ are special cases where $g$ is injective and $f$ is surjective, respectively.
That's it and the link to Wikipedia in the first comment includes this and more.