The following is taken from $\textit{Module Theory An Approach to Linear Algebra}$ By: T.S.Blyth
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$\textbf{Exercise 4.5}$ $\textbf{[The snake diagram]}$ Suppose that the diagram of $R-$modules and $R-$morphisms
(1)
is commutative and has exact rows. Show that this diagram can be extended to a diagram
(2)
which is also commutative and has exact rows and columns. Show also that there isa 'connecting $R-$morphism' $d:\text{Ker }\gamma\to A'/\text{im }\alpha$ such that
$\text{ker }\alpha\xrightarrow{u_1}\text{ker }\beta\xrightarrow{v_1}\text{ker }\gamma\xrightarrow{d}A'/\text{Im }\alpha\xrightarrow{u_2}B'/\text{Im }\beta\xrightarrow{v_2}C'/\text{Im }\gamma$
is exact.
[$\textit{Hint.}$ To construct $d:$ given $x\in\text{Ker }\gamma$ let $y\in B$ be such that $v(y)=k(x).$ Show that $\beta(y)\in \text{ker }v'$ so that there exists a unique $a'\in A'$ such that $u'(a')=\beta(y).$ Show that the prescription $d(x)=p(a')$ is well defined (i.e., independent of $\gamma$) and does the trick.]
$\color{Red}{Questions:}$
I would like to know for the exercise above, what does it when it ask the reader to "Show that this diagram can be extended to a diagram"? Does it mean that I can build another commutative diagram from (1) so that the following exact sequence
$\text{ker }\alpha\xrightarrow{u_1}\text{ker }\beta\xrightarrow{v_1}\text{ker }\gamma$
is exact? Similarly, I can build another bottom commutative diagram connecting (1) to show the following exact sequence
$A'/\text{Im }\alpha\xrightarrow{u_2}B'/\text{Im }\beta\xrightarrow{v_2}C'/\text{Im }\gamma$
is exact?
Thank you in advance