I give an elementary proof of the fact that a quasi-isomorphism of short complexes (complexes of length 2) of $\Gamma$-modules induces an isomorphism on hypercohomology.
Actually, it is very close to Theorem 3.3 on page 16 of
my preprint of 1992,
where I prove that a quasi-isomorphism of crossed modules with $\Gamma$-action preserves ${\Bbb H}^0$ and ${\Bbb H}^1$ (for any profinite group $\Gamma$).
First I give my elementary cocyclic definitions of cohomology and hypercohomology.
Definition of cohomology
Let $A$ be a $\Gamma$-module, that is, an abelian group written additively, endowed with an action of $\Gamma=\{1,\gamma\}$.
We consider the first cohomology group $H^1(\Gamma,A)$.
We write $H^1\! A$ for $H^1(\Gamma,A)$.
Recall that
$$H^1\! A=Z^1\! A/B^1\! A,$$
where
$$Z^1\! A=\{a\in A\mid\,^\gamma\! a=-a\}, \quad
B^1\! A=\{\,^\gamma\! a'-a'\mid a'\in A\}.$$
We define the second cohomology group $H^2\!A$ by
$$ H^2\!A=Z^2\! A/ B^2\! A,$$
where
$$Z^2\! A=A^\Gamma:= \{a\in A\mid\, ^\gamma\! a=a\},\quad\
B^2\! A=\{\,^\gamma\! a'+a'\mid a'\in A\}.$$
For $k\in{\mathbb Z}$ we define the coboundary operator
$$d^k\colon A\to A,\quad a\mapsto\,^\gamma\! a-(-1)^k a.$$
In other words, $d^k=\gamma-(-1)^k\in {\mathbb Z}[\Gamma]$, where ${\mathbb Z}[\Gamma]={\mathbb Z}\oplus{\mathbb Z}\gamma$ is the group ring of $\Gamma$.
We calculate:
\begin{align*}
d^k\circ d^{k-1}=\big(\gamma-(-1)^k\big)\big(\gamma-(-1)^{k-1}\big)&=\big(\gamma-(-1)^k\big)\big(\gamma+(-1)^k\big)\\
&=\gamma^2-(-1)^{2k}=1-1=0.
\end{align*}
We see that $d^{k+1}\circ d^k=0$.
We define the Tate cohomology groups $H^k_T A$ for all $k\in {\mathbb Z}$ by
$$ H^k_T\, A=Z^k\! A/B^k\! A,$$
where
\begin{align*}
Z^k\! A=\ker d^k=\{a\in A\mid\,^\gamma\! a=(-1)^k a\},\quad
B^k\! A={\rm im\,} d^{k-1}=\{\,^\gamma\! a'+(-1)^k a'\mid a'\in A\}.
\end{align*}
Then clearly
$$H^k_T\, A=H^1\! A\quad\text{if $k$ is odd, $\ $ and}\quad H^k_T\, A=H^2\!A\ \ \text{if $k$ is even.}$$
Definition of hypercohomology
Let $A_0\overset{\partial}{\longrightarrow} A_0$ be a short complex of $\Gamma$-modules.
For $k\in{\mathbb Z}$ we define a differential
$$D^k\colon\, A_1\oplus A_0\to A_1\oplus A_0,\quad\ D(a_1,a_0)=\big(d^{k+1} a_1,\, d^k a_0-(-1)^k a_1\big).$$
We calculate:
\begin{align*}
D^k\big( D^{k-1}(a_1,a_0)\big)&=D^k\big(\, d^k a_1,\ d^{k-1} a_0-(-1)^{k-1} a_1\,\big)\\
&=\big(d^{k+1}d^k a_1,\ d^k d^{k-1} a_0-(-1)^{k-1}d^k a_1- (-1)^k d^k a_1\big)=0.
\end{align*}
Thus $D^k\circ D^{k-1}=0$.
We define the $k$-th Tate hypercohomology group ${\mathbb H}^k(A_{1}{\overset\partial\longrightarrow } A_0)$ by
$$ {\mathbb H}^k(A_{1}{\overset\partial\longrightarrow } A_0)=Z^k(A_{1}{\overset\partial\longrightarrow } A_0)\,/\,B^k(A_{1}{\overset\partial\longrightarrow } A_0),$$
where
\begin{align*}
Z^k(A_{1}&{\overset\partial\longrightarrow } A_0)=\ker D^k=\\
&=\{(a_{1},a_0)\in A_{1}\oplus A_0,\,\, \mid\,d^{k+1} a_1=0,\, d^k a_0-(-1)^k\partial a_{1}=0\}, \\
B^k(A_{1}&{\overset\partial\longrightarrow } A_0)={\rm im\,} D^{k-1}=\\
&=\{(d^k a'_1,\ d^{k-1} a'_0-(-1)^{k-1}\partial a'_1\,) \,\mid\, (a'_1, a_0')\in A_1\oplus A_0\}.
\end{align*}
Quasi-isomorphism induces isomorphism on hypercohomology
Proposition. If $\varphi=(\varphi_1,\varphi_0)\colon\, (A_1\overset{\partial}{\longrightarrow} A_0)\,\longrightarrow\,
(A_1'\overset{\partial'}{\longrightarrow} A_0')$ is a quasi-isomorphism, then the induced homomorphism on hypercohomology
$$\varphi_*\colon\ {\Bbb H}^k(\Gamma, A_1\to A_0)\,\to\, {\Bbb H}^k(\Gamma, A_1'\to A_0')$$
is an isomorphism.
Sketch of an elementary proof.
From $A_1\overset{\partial}{\longrightarrow} A_0$, using the short exact sequence of complexes
$$0\to(\ker\partial\to 0)\to (A_{1}\overset\partial\longrightarrow A_0)\to ({\rm im}\, \partial\hookrightarrow A_0)\to 0,$$
we obtain a long exact sequence
$$\dots H^{k-1}{\rm coker}\,\partial\to H^{k+1}\ker\partial\to {\mathbb H}^k(A_{1}\overset\partial\longrightarrow A_0)
\to H^k{\rm coker}\,\partial\to H^{k+2}\ker\partial\dots, $$
and similarly, from $A_1'\overset{\partial'}{\longrightarrow} A_0'$ we obtain a long exact sequence
$$\dots H^{k-1}{\rm coker}\,\partial'\!\to\! H^{k+1}\ker\partial'\!\to\! {\mathbb H}^k(A_{1}'\!\overset{\partial'}\longrightarrow\! A_0')
\!\to\! H^k{\rm coker}\,\partial'\!\to H^{k+2}\ker\partial'\!\dots $$
Our morphism $\varphi=(\varphi_1,\varphi_0)$ induces a morphism of the exact sequences,
that is, a commutative diagram with exact rows.
Since our morphism $\varphi$ is a quasi-isomorphism, the induced homomorphisms
$$ H^{k}\ker\partial\to H^{k}\ker\partial'\quad\text{and}
\quad H^k{\rm coker}\,\partial\to H^k{\rm coker}\,\partial' $$
are isomorphisms for all $k$, and by the five lemma we conclude that
the induced homomorphism on hypercohomology
$$\varphi_*\colon\ {\Bbb H}^k(\Gamma, A_1\to A_0)\,\to\, {\Bbb H}^k(\Gamma, A_1'\to A_0')$$
is an isomorphism, as required.
