TITLE: What is the meaning of this coboundary homomorphism for group hypercohomology?
QUESTION [5 upvotes]: $\require{AMScd}$
Let $\Gamma=\{1,\gamma\}$ be a group of order 2.
In my problem from Galois cohomology of real reductive groups I came to a commutative diagram of $\Gamma$-modules
(abelian groups with $\Gamma$-action)
\begin{equation*}%\label{e:cd}
\begin{CD}
1 @>>>Q_1 @>>>Q_2 @>>>  Q_3     @>>>  1 \\
@.  @VV{\rho_1}V @VV{\rho_2}V   @VV{\rho_3}V \\
1 @>>>  X_1 @>>> X_2 @>>>   X_3 @>>>  1 \\
@.  @VV{\alpha_1}V @VV{\alpha_2}V@VV{\alpha_3}V \\
1 @>>>  P_1 @>>> P_2 @>>>    P_3 @>>>  1 \\
\end{CD}
\end{equation*}
in which the rows are exact, but not the columns
(and $\alpha_k\circ\rho_k\neq 0$).
The top and bottom rows of the diagram split canonically:
$$Q_2=Q_1\oplus Q_3\quad\text{ and }\quad P_2=P_1\oplus P_3,$$
and these splittings are compatible:
$$
\alpha_2(\rho_2(0,q_3))=
\big(\,0,\,\alpha_3(\rho_3(q_3))\,\big)\tag{$*$}
$$
for $q_3\in Q_3$.
I consider the Tate hypercohomology groups
$${\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)\quad\text{ and }
           \quad{\Bbb H}^0(\Gamma,X _1\overset{\alpha_1}\longrightarrow P_1),$$
where both short complexes are in degrees $(-1,0)$.
Below I construct "by hand" a canonical coboundary homomorphism
$$\delta\colon\, {\Bbb H}^0(\Gamma, Q_3\to X _3)\,\longrightarrow\, 
{\Bbb H}^0(\Gamma,X _1\to P_1),$$

Question. How can I get this coboundary homomorphism from a kind of general theory?

Remark. For a group $\Gamma$ of order 2 (and also for any cyclic group $\Gamma$) the Tate cohomology and hypercohomology are periodic with period 2.
Therefore, our $\delta$ is a map
$${\Bbb H}^1(\Gamma,\, Q_3\to X_3\to 0)\, \longrightarrow \, 
{\Bbb H}^2(\Gamma,\, 0\to X_1\to P_1),$$
where both complexes are in degrees $(-2,-1,0)$.
Construction.
We start with $[ q_3,  x_3]\in {\Bbb H}^0(\Gamma, Q_3\overset{\rho_3}\longrightarrow X _3)$.
Here $( q_3,  x_3)\in  Z^0(\Gamma,Q_3\to X _3)$, that is,
\begin{equation}
q_3\in Q_3,\quad x_3\in X_3,\quad
\,^{\gamma\kern -0.8pt} q_3+ q_3=0,\qquad \,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3).\tag{$**$}
\end{equation}
We lift canonically $ q_3$ to
$$ q_2=(0, q_3)\in Q_1\oplus Q_3= Q_2,$$
and we lift $ x_3$ to some $ x_2\in X _2$.
We write
$$\alpha_2( x_2)=( p_1, p_3)\in P_1\oplus P_3=P_2,$$
where $ p_3=\alpha_3( x_3)\in P_3$ and $ p_1\in P_1$.
We set
$$ x_1=\,^{\gamma\kern -0.8pt} x_2- x_2-\rho_2( q_2).$$
Since by $(*)$  we have
$$\,^{\gamma\kern -0.8pt} x_3- x_3=\rho_3( q_3),$$
we see that  $ x_1\in X _1$.
We compute:
$$\,^{\gamma\kern -0.8pt} x_1+ x_1=\,^{\gamma\kern -0.8pt}(\,^{\gamma\kern -0.8pt} x_2- x_2)-{}^{\gamma\kern -0.8pt}\rho_2(0, q_3)+
    (\,^{\gamma\kern -0.8pt} x_2- x_2)-\rho_2(0, q_2)=-\rho_2(0,\,^{\gamma\kern -0.8pt} q_3+ q_3)=0$$
by $(**)$.
Furthermore,
\begin{align*}
\alpha_1( x_1)&=\,^{\gamma\kern -0.8pt}\alpha_2(x_2)-\alpha_2(x_2)-\alpha_2(\rho_2(q_2))\\
&=\,^{\gamma\kern -0.8pt}( p_1, p_3)-( p_1, p_3)-( 0,\alpha_3(\rho_3( q_3)))\\
&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,^{\gamma\kern -0.8pt}p_3-p_3-\alpha_3(\rho_3(q_3))\big)\\
&=\big(\,^{\gamma\kern -0.8pt}p_1-p_1,\,\alpha_3(\,^{\gamma\kern -0.8pt}x_3-x_3-\rho_3(q_3))\big)\\
&=(\,^{\gamma\kern -0.8pt} p_1- p_1,0)
\end{align*}
by $(*)$ and $(**)$.
Thus
$$\alpha_1(x_1)=\,^{\gamma\kern -0.8pt} p_1-p_1.$$
We see that $(x_1, p_1)\in Z^0(\Gamma, X _1\overset{\alpha_1}\longrightarrow P_1)$.
We set
$$\delta[ q_3, x_3]=[ x_1, p_1]\in {\Bbb H}^0(\Gamma,X _1\to P_1).$$
A straightforward check shows that the map $\delta$ is a well-defined homomorphism.

REPLY [3 votes]: I believe easiest way to handle this is in the formalism of triangulated categories. You can do it in various ways: either work with the unbounded derived category or (probably easier) replace each module $M$ with $\operatorname{Hom}_\Gamma(\mathcal R,M)$ where $\mathcal R$ is the complete resolution for $\Gamma$, i. e. the standard unbounded 2-periodic complex $$\cdots\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}\mathbb Z[\Gamma]\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}\cdots$$of $\Gamma$-modules.
Let then $X_1\to X_2\to X_3\to\Sigma X_1$ be an exact triangle in arbitrary triangulated category, and let $Q_3\to X_2\to P_1$ be arbitrary morphisms with zero composite. Let $P$ be the fiber of $X_1\to P_1$ and let $Q$ be the cofiber of $Q_3\to X_3$. Our aim is to construct from all that a canonical map $Q\to\Sigma P$. It turns out that there is such a map which is moreover an isomorphism if and only if $Q_3\to X_2\to P_1$ is exact.
Since the composite $Q_3\to X_2\to P_1$ is zero, the map $X_2\to P_1$ factors through cofiber of $Q_3\to X_2$, $X_2\to Q_0$, and the map $Q_3\to X_2$ factors through the fiber $P_0\to X_2$ of $X_2\to P_1$. Thus all in all $X_1\to P_1$ factors into the composite $X_1\to X_2\to Q_0\to P_1$, while $Q_3\to X_3$ factors into the composite $Q_3\to P_0\to X_2\to X_3$.
First note that in these circumstances the cofiber of $Q_3\to P_0$ is isomorphic to the fiber of $Q_0\to P_1$; denoting it by $H$, the composite $P_0\to H\to Q_0$ is the composite $P_0\to X_2\to Q_0$.
We get eight instances of the octahedron axiom, telling us that for various composites $f\circ g$ there are exact triangles $\operatorname{fibre}(f)\to\operatorname{cofibre}(g)\to\operatorname{cofibre}(f\circ g)\to\operatorname{cofibre}(f)=\Sigma\operatorname{fibre}(f)$ and $\operatorname{fibre}(g)\to\operatorname{fibre}(f\circ g)\to\operatorname{fibre}(f)\to\operatorname{cofibre}(g)=\Sigma\operatorname{fibre}(g)$. Strictly speaking, not all of them are needed, but for completeness let me list them all.




The composable pair
gives the exact triangle




$Q_3\to P_0\to X_2$
$H\to Q_0\to P_1\to\Sigma H$


$Q_3\to X_2\to X_3$
$X_1\to Q_0\to Q\to \Sigma X_1$


$Q_3\to P_0\to X_3$
$\color{red}{P\to H\to Q\to\Sigma P}$


$P_0\to X_2\to X_3$
$P\to X_1\to P_1\to\Sigma P$


$X_1\to X_2\to Q_0$
$Q_3\to X_3\to Q\to\Sigma Q_3$


$X_1\to X_2\to P_1$
$P\to P_0\to X_3\to\Sigma P$


$X_1\to Q_0\to P_1$
$\color{red}{P\to H\to Q\to\Sigma P}$


$X_2\to Q_0\to P_1$
$Q_3\to P_0\to H\to\Sigma Q_3$




To put it all in a single diagram - in what follows, lines with three objects on them represent exact triangles; everything commutes.