I'm attempting to answer the following:
In a commutative ladder of R-modules with exact rows:
where $\gamma_n$ are isomorphisms. Prove that sequence:
$$ \dots \rightarrow A_n \overset{(\alpha_n,-i_n)}\longrightarrow A'_n\oplus B_n \overset{\langle i'_n,\beta_n\rangle}\longrightarrow B'_n \overset{(\delta_n\gamma_n^{-1}j'_n)}\longrightarrow A_{n_1} \to \dots$$
is exact, where $\langle i'_n,\beta_n\rangle \colon (a',b)\mapsto i'_n(a')+\beta_n(b)$.
We need to show exactness at three place, e.g. at $A'_n \bigoplus B_n$ it is: $i'_n(a')+\beta_n(b)=0\in B$ iff $\exists a_n\in A_n. (a',b)=(\alpha_n(a),-i_n(a))$. From commutativity we get: $\beta_n\circ i_n= i'_n \circ \alpha_n$.
At $B'_n$: $\exists a',b.i_n(a')+\beta_n(b)=b'$ iff $\partial_n\gamma_n^{-1}j'_n(b')=0\in A_{n-1}$. Isn't the right condition always true, because every element of exact sequence vanishes after mapped twice (taking into account $\gamma_n$ is iso).
Shall I use some prepared result such as five or snake lemma?
I know my answer is a bit late but I just stumbled across the same problem, I won‘t explicitly write all the indexes, I hope it’s still clear:
The „the image is contained in the kernel“ inclusions are easy:
„$\text{Im}(\alpha,-i) \subseteq \ker<i‘, \beta>$“: $i‘\circ \alpha- \beta \circ i = 0$ by commutativity.
„$\text{Im} <i‘, \beta> \subseteq \ker \delta \circ \gamma^{-1} \circ j‘$“: $\delta \circ \gamma^{-1} \circ j‘ \circ(i‘+\beta)= \delta \circ \gamma^{-1} \circ j‘ \circ \beta $ because $j‘\circ i‘=0$ and because $j‘\circ \beta = \gamma \circ j$ we get $\delta \circ \gamma^{-1} \circ j‘ \circ \beta =\delta \circ j=0$.
„$\text{Im} \delta \circ \gamma^{-1} \circ j‘ \subseteq \ker (\alpha,-i) $“: Follows because $\alpha \circ \delta \circ \gamma^{-1} \circ j‘ =\delta‘ \circ j‘=0$ and $-i \circ \delta \circ \gamma^{-1} \circ j‘=0$ (since $i \circ \delta =0$)
The other three inclusions are a bit harder:
„$\ker <i‘, \beta> \subseteq \text{im} (\alpha,-i)$“: Let $i‘(a‘)+\beta(b)=0$ then $j‘\circ \beta(b)=0$ thus $j(b)=0$ and there exists an $a\in A$ s.t. $i(a)=b$. Thus $i‘ \circ \alpha(a)=\beta(b)=-i‘(a‘)$ hence there is a $c‘ \in C‘$ s.t. $\delta‘(c)=\alpha(a)+a‘$. Now we have everything we need: $$ (\alpha(\delta\circ \gamma^{-1}(c)-a), -i(\delta\circ \gamma^{-1}(c)-a))=(\delta‘\circ\gamma \circ \gamma^{-1}(c)-\alpha(a), i(a))=(a‘,b) $$
„$\ker \delta \circ \gamma^{-1} \circ j‘ \subseteq \text{im} <i‘, \beta> $“: Let $\delta \circ \gamma^{-1} \circ j‘ (b‘)=0$ then there is a $b \in B$ s.t. $j(b)=\gamma^{-1} \circ j‘(b‘)$ but this means that $j‘(b‘-\beta(b))=\gamma \circ j(b)-j‘ \circ \beta(b)=0$ so there also is an $a‘ \in A‘$ s.t. $i‘(a‘)=b‘-\beta(b)$ which means: $$ <i‘, \beta>(a‘,b)=b‘-\beta(b)+\beta(b)=b‘ $$
„$\ker (\alpha,-i) \subseteq \text{im} \delta \circ \gamma^{-1} \circ j‘ $“: Let $\alpha(a)=0=i(a)$ thus there is a $c \in C$ s.t. $\delta(c)=a$ and since $\delta‘ \circ \gamma(c)=\alpha \circ \delta(c)=0$ there also is a $b‘ \in B‘$ s.t. $j‘(b‘)=\gamma(c)$ all this means:
$$ \delta \circ \gamma^{-1} \circ j‘(b‘)=\delta(c)=a $$