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.]
$\color{Red}{Questions:}$
In this youtube video: Snake Lemma (a VERY COMPLETE proof) at the 7:14 mark, the presenter talked about showing the maps $i,j$ are well defined.
But in this post on a proof about deriving $\text{ker }(\alpha) \xrightarrow{f|_{\text{ker }(\alpha)}} \text{ker }(\beta),$
[in that post, $a=\alpha, b=\beta, u_1={f\mid_{\text{ker }(\alpha)},}$ and $u_i$ corresponds to map $i$ in the youtube video above.] I inquired about the details how the universal property of the kernel is used in the quoted proof. That proof did not say anything about the need to show the map $f|_{\text{ker }(\alpha)}$ is well defined. I am just wondering if the sequence of maps $\text{ker }(\alpha) \xrightarrow{u_1} \text{ker }(\beta)\xrightarrow{v_1}\text{ker }(\gamma)$ need to be shown it is well defined, or is it enough when showing their constructions, simply use the universal mapping property for kernel, and that will be enough.
Thank you in advance