This exercise shows that the Snake Lemma implies Theorem $6.10$ (so this theorem should not be used in solving this problem).
Consider the commutative diagram with exact rows (note that two zeros are "missing" from this diagram):
(i) Prove that $\Delta:$ ker $\gamma \rightarrow \operatorname{coker} \alpha$, defined by
$\Delta: z \mapsto i^{-1} \beta p^{-1} z+i m \alpha$,
is a well-defined homomorphism.
(ii) Prove that there is an exact sequence
$\operatorname{ker} \alpha-\operatorname{ker} \beta \rightarrow \operatorname{ker} \gamma \stackrel{\Delta}{\rightarrow} \operatorname{coker} \alpha \stackrel{i^{r}}{-} \operatorname{coker} \beta \rightarrow \operatorname{coker} \gamma$,
where $i^{\prime}: a^{\prime}+\operatorname{im} \alpha \mapsto i a^{\prime}+\operatorname{im} \beta$ for $a^{\prime} \in A^{\prime}$.
(iii) Given a commutative diagram with exact rows,
$$
0 \longrightarrow A_{n}^{\prime} \longrightarrow A_{n} \longrightarrow A_{n}^{n} \longrightarrow 0
$$
prove that the following diagram is commutative and has exact rows:
$$
\begin{gathered}
A_{n}^{r} / \operatorname{im} d_{n+1}^{\prime} \rightarrow A_{n} / \operatorname{im} d_{n+1} \rightarrow A_{n}^{\prime \prime} / \operatorname{im} d_{n+1}^{\prime \prime} \rightarrow 0 \\
d^{\prime} \downarrow_{k e r} d_{n-1}^{\prime} \longrightarrow \operatorname{ker} d_{n-1}^{\prime} \longrightarrow \operatorname{ker} d_{n-1^{*}}^{\prime \prime}
\end{gathered}
$$
(iv) Use part (ii) and this last diagram to give another proof of Theorem 6.10, the Long Exact Sequence.