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$\require{AMScd}$ 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.

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    $\begingroup$ I don't see the details but your data, I believe, reduce to arbitrary homomorphisms $Q_3\to X_2\to P_1$. Do $Q_1\to X_1$ or $X_3\to P_3$ play any role? In any case I believe the spectral sequences relevant for comparing hypercohomologies of $X_1\to X_2$ and $X_2\to P_1$ with those of $X_1\to P_1$, of $Q_3\to X_2$ and $X_2\to P_1$ with those of $Q_3\to P_1$, and of $Q_3\to X_2$ and $X_2\to X_3$ with those of $Q_3\to X_3$ should reduce to interleaved four-tuples of long exact sequences which would then include your map. $\endgroup$ Jan 10, 2021 at 22:00
  • $\begingroup$ @მამუკა ჯიბლაძე: Thank you, this was helpful. Indeed, the homomorphisms $Q_1\to X_1$ and $X_3\to P_3$ play no role. The composite homomorphism $Q_3\to X_2\to P_1$ is zero: I have added condition $(*)$. $\endgroup$ Jan 11, 2021 at 10:42
  • $\begingroup$ But $Q_3\to X_2\to P_1$ is not exact? $\endgroup$ Jan 11, 2021 at 10:53
  • $\begingroup$ No, this sequence is not exact. $\endgroup$ Jan 11, 2021 at 11:29
  • $\begingroup$ Actually, in my application all groups are finitely generated free abelian groups. Moreover, ${\rm rk}\,P_k={\rm rk}\, Q_k$ for $k=1,2,3$, but I don't use it. The groups $X_k$ may have big rank, independent of the rank of $P_k$ and $Q_k$. $\endgroup$ Jan 11, 2021 at 11:33

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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.

enter image description here

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