$\def\Ker{\operatorname{Ker}} \def\Im{\operatorname{Im}}$Given an exact couple $(A,E,\alpha,f,g)$ in some abelian category, we define its derived exact couple $(A',E',\alpha',f',g')$ to be $A'=\alpha A=\operatorname{Im}\alpha$, $E=\Ker d/\Im d$ where $d=g\circ f$, the map $\alpha':A'\to A'$ is restriction of $\alpha$ to $\Im \alpha$ and the maps $f'$, $g'$ are the connecting homomorphism respectively obtained by application of the snake lemma to the following diagrams with exact rows: $$ \require{AMScd} \begin{CD} @.E@>{d}>> E@>>>E/\Im d@>>> 0\\ @.@V{0}VV@V{f}VV@VV{d}V@.\\ 0@>>>A/\Ker \alpha@>\smash\alpha>> A@>\smash g>> E \end{CD} $$ and $$ \require{AMScd} \begin{CD} @.E@>{f}>> A@>{\alpha}>>\alpha A@>>> 0\\ @.@V{d}VV@V{g}VV@VV{0}V@.\\ 0@>>>\Ker d@>>> E@>\smash d>> E \end{CD} $$ Note that the sequences obtained from the snake lemma give exactness of $$ \label{lseq}\tag{1} E'\xrightarrow{f'}A'\xrightarrow{\alpha'}A'\xrightarrow{g'}E'. $$ Exactness at the domain (resp., codomain) of $\alpha$ comes from the first (resp., second) diagram. Thus, to verify that $(A',E',\alpha',f',g')$ is indeed an exact couple, it is left to check that $$ \label{seq}\tag{2} A'\xrightarrow{g'}E'\xrightarrow{f'}A' $$ is exact. Every source I've checked leaves this as an exercise to the reader. In fact, no source I've read defines $f',g'$ the way I've done, and instead the definition of these maps is hinted like in 011R.
Of course I can resort to $R$-modules and do a good ol' diagram-chasing proof to check that \eqref{seq} is exact. My concern is:
Can we prove exactness of \eqref{seq} from some of the diagram lemmas? (Snake, salamander, $n\times n$, four, five...) Or any indirect method that does not require chasing elements or morphisms.
Even though I achieved this with \eqref{lseq}, I tried with \eqref{seq} but I cannot find the right idea.
