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.
