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 $...\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}\mathbb Z[\Gamma]\xrightarrow{1-\gamma}\mathbb Z[\Gamma]\xrightarrow{1+\gamma}...$ 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][1]][1]


  [1]: https://i.sstatic.net/ZFYdH.png