$\newcommand{\Ext}{\operatorname{Ext}^1}$I am studying the Hilton & Rees article on natural transformations between $\Ext$ functors which can be found here.
More specifically, in Theorem 1.3 they prove that every natural transformation $\varphi:\Ext(B,-)\to \Ext(A,-)$ is induced by a homomorphism $\eta\in \hom(A,B)$.
They use the following result:
Let $\require{AMScd}\begin{CD} 0 @>{}>>A^{\prime\prime} @>{}>> A^\prime @>{}>> A @>{}>> 0 \end{CD}$ be a short exact sequence of modules.
We have connecting homomorphisms $\require{AMScd}\begin{CD} \ldots @>{}>> \hom(A^{\prime\prime},A^{\prime\prime}) @>{\partial}>> \Ext(A,A^{\prime\prime}) @>{}>>\ldots \end{CD}$
and
$\require{AMScd}\begin{CD} \ldots @>{}>>\hom(A,A) @>{\delta}>> \Ext(A,A^{\prime\prime}) @>{}>>\ldots \end{CD}$
Then $\partial(id_{A^{\prime\prime}})+\delta(id_A)=0$
I can't prove this result. The article references Cartier's article which can be found here. I had a very hard time studying the latter. I couldn't find the proof and my French is really weak.
Any help with the proof of this result would be greatly appreciated. Thank you.
Add If you have set up the machinery of suspended (or triangulated) categories, it seems the proof is the following: your SES is a morphism $\xi : A'\to A''[1]$, and $\partial(1_{A''}) = 1_{A''}[1]\circ \xi = -\xi$ since translating the identity introduces a sign, while $\delta(1_{A'}) = \xi \circ 1_{A'} = \xi$.
$\newcommand{\Ext}{\operatorname{Ext}^1}$Let me write your exact sequence $S$ and fix another module $B$, so that you want to describe two connecting morphisms
$$\partial:\hom_R(A'',B)\to \Ext_R(A',B), \qquad\delta:\hom_R(B,A')\to \Ext_R(A',A'').$$ So take your SEC $0\to A''\to A\to A'\to 0$. You want to describe the following connecting maps somehow and show that $\partial 1_{A''}+\delta 1_{A'}=0$. One way to proceed is as follows. Take a map $f:A''\to B$, and consider the diagram
\begin{CD} 0 @>>>B @>>> E @>>> A'@>>>0\\ {}&&@AfAA @AAA @AA{1_A'}A \\ 0@>>>A'' @>>> A @>>> A'@>>>0 \\ \end{CD}
where the first square is a pushout. Then the class of the extension in the first row is $\partial f$. Similarly, if you take $g:B\to A'$ and consider the diagram where the second square is a pullback, then the class of the extension in the second row is $\delta g$.
\begin{CD} 0 @>>>A'' @>>> A @>>> A'@>>>0\\ {}&&@A{1_{A''}}AA @AAA @AAgA \\ 0@>>>A'' @>>> F @>>> A'@>>>0 \\ \end{CD}
We now consider your case, where the SES $S$ goes (almost) to itself. I claim that the extension $\partial 1_{A''}$ is isomorphic to the extension
$$0\to A''\to A\to A'\to 0$$
where the first arrow, say $f$, now has a minus sign, that is, it is $-f$. Indeed, when you consider the pushout, $E = A''\oplus A /\langle (a'',-fa'')\rangle$ and the induced map $A''\to E \simeq A$ is given by $$a''\to \textrm{class}(a'',0) = \textrm{class}(0,-fa'') \simeq -fa''\in A,$$
while the projection $E\to E/\langle (0,-fa'')\rangle = \operatorname{coker} f \simeq A$ identifies with $g$, without signs.
On the other hand, when doing the pullback, no signs are involved. To conclude, the sum of two extensions represented by SESs is given by the Baer sum, and it is a fact that negating one of the morphisms in a SES gives you its additive inverse for the Baer sum.