$\DeclareMathOperator{\Tot}{Tot}$I have the following problem. Let $H$ and $G$ be groups such that $H$ acts on $G$, i.e., there exists a group homomorphism $H\to \mathrm{Aut}(G)$ and let $M$ be an abelian group treated as a trivial $G$- and $H$-module.


Further on, there is a double complex $C^{p,q}:= C^p(H,C^q(G,M))$ (I use here bar resolutions). To this double complex there are two spectral sequences associated, one of them is Lyndon--Hochschild--Serre spectral sequence. In particular, I know that both spectral sequences converge to $H^\ast(G\rtimes H, M)$. On the other hand, by the general construction of both spectral sequences, we know that they converge to $H^\ast(\Tot(C^{\ast \ast}))$.


My question is the following. If I have a specific cohomology class in $H^\ast(\Tot(C^{\ast \ast}))$, how can I decide to which class in $H^\ast(G\rtimes H, M)$ it corresponds? In particular, an answer to any of the following questions would do the job (I think):


1. Is there an explicit map $H^\ast(\Tot(C^{\ast \ast}))\to H^\ast(G\rtimes H, M)$ giving the isomorphism? Or is there a cochain level map?
2. Given a class in $H^2(\Tot(C^{\ast \ast}))$, how can I construct a group extension $1\to M \to E\to G\rtimes H\to 1$?


I am mostly interested in cohomology classes of degree 2 (as indicated by the second question). Also, reference would be enough.



Comparing cohomology of a total complex with the cohomology of semidirect product
Let me change the notation a little: let $\phi \colon H \to \operatorname{Aut}(N)$ be a group homomorphism, and consider $G = N \rtimes\_\phi H$ (we will drop the subscript $\phi$ in the rest of this post). As usual, write $^hn$ for $\phi(h)(n)$ if $h \in H$ and $n\in N$.


The point is that the double complex $C^{\bullet\bullet}$ actually comes from a natural bi-cosimplicial abelian group, as follows:


**Definition.** Define the bisimplical set $X\_{\bullet\bullet}$ via $X\_{p,q} = N^{q+1}\times H^{p+1}$, where a morphism $(f,g) \colon ([p],[q]) \to ([p'],[q'])$ in $\Delta \times \Delta$ acts as
\begin{align\*}
X\_{p',q'} &\to X\_{p,q}\\
\big(n\_0,\ldots,n\_{q'},h\_0,\ldots,h\_{p'}\big) &\mapsto \big(n\_{f(0)},\ldots,n\_{f(q)},h\_{g(0)},\ldots,h\_{g(p)}\big).
\end{align\*}
Endow each $X\_{p,q}$ with a $G$-action via
\begin{align\*}
G \times X\_{p,q} &\mapsto X\_{p,q}\\
\big((n,h),(n\_0,\ldots,n\_q,h\_0,\ldots,h\_p)\big) &\mapsto \big(n\ {^hn\_0},\ldots,n\ {^hn\_q},hh\_0,\ldots,hh\_p\big)\\
\big((n,h),\big(\underline n,\underline h\big)\big) &\mapsto \big(n\ ^h\underline n,h\underline h\big).
\end{align\*}
This is visibly functorial in $(p,q) \in \Delta^{\text{op}} \times \Delta^{\text{op}}$, so defines a bisimplicial $G$-set. Thus we can view $\mathbf Z[X\_{\bullet\bullet}]$ naturally as a bisimplicial $\mathbf Z[G]$-module, and I claim that
$$\operatorname{Hom}\_{\mathbf Z[G]}\big(\mathbf Z[X\_{\bullet\bullet}],M\big) \cong C^{\bullet\bullet}\label{1}\tag{1}$$
as double complexes, for any $G$-module $M$ (not necessarily trivial). Indeed, note that
\begin{align\*}
\operatorname{Hom}\_{\mathbf Z[G]}\big(\mathbf Z[X\_{p,q}],M\big) &\cong \operatorname{Map}\_G\big(X\_{p,q},M\big), \\
C^{p,q} &\cong \operatorname{Map}\_H\big(H^{p+1},\operatorname{Map}\_N(N^{q+1},M)\big).
\end{align\*}
We have a bijection $\operatorname{Map}(X\_{p,q},M) \stackrel\sim\to \operatorname{Map}(H^{p+1},\operatorname{Map}(N^{q+1},M))$, taking a function $\phi \colon X\_{p,q} \to M$ to $\psi \colon H^{p+1} \to \operatorname{Map}(N^{q+1},M)$ given by
$$\psi\big(h\_0,\ldots,h\_p\big)\big(n\_0,\ldots,n\_q\big) = \phi\big(n\_0,\ldots,n\_q,h\_0,\ldots,h\_p\big).$$
Now $\psi$ has image in $\operatorname{Map}\_N(N^{q+1},M)$ if and only if
$$\phi\big(n\underline n,\underline h\big) = n\phi\big(\underline n,\underline h\big) \qquad \text{for all } \big(\underline n, \underline h\big) \in X\_{p,q}, \text{ all } n \in N.\label{2}\tag{2}$$
The $G$-action on $\operatorname{Map}\_N(N^{q+1},M)$ is given by $(gf)(\underline n) = gf(g^{-1} \cdot \underline n)$, which is trivial on $N$ by definition of $\operatorname{Map}\_N$. As action of $H$, it can be written explicitly by $(hf)(\underline n) = hf(^{h^{-1}}\!\underline n)$. This means that $\psi \colon H^{p+1} \to \operatorname{Map}\_N(N^{q+1},M)$ is $H$-equivariant if and only if
$$\phi\big(\underline n,h\underline h\big) = h\phi\big({^{h^{-1}}}\!\underline n,\underline h\big) \qquad \text{for all } \big(\underline n,\underline h\big) \in X\_{p,q}, \text{ all } h \in H.\label{3}\tag{3}$$
Combining \eqref{2} and \eqref{3} shows that $\psi \in C^{p,q}$ if and only if $\phi(g \cdot (\underline n,\underline h)) = g \phi(\underline n,\underline h)$ for all $(\underline n,\underline h) \in X\_{p,q}$ and all $g$ in either $N$ or $H$. But this means exactly that $\phi \colon X\_{p,q} \to M$ is $G$-equivariant, so
$$\operatorname{Map}\_G\big(X\_{p,q},M\big) \stackrel\sim\to \operatorname{Map}\_H\big(H^{p+1},\operatorname{Map}\_N\big(N^{q+1},M\big)\big).$$
This identification is natural in $p$ and $q$, giving the isomorphism \eqref{1}. This also shows that $C^{\bullet\bullet}$ is naturally a bi-cosimplicial abelian group.


In particular, the Eilenberg–Zilber theorem ([nlab](https://ncatlab.org/nlab/show/Eilenberg-Zilber%20theorem#statement)) shows that the totalisation of $C^{\bullet\bullet}$ is naturally chain homotopic (!) to the diagonal $dC^\bullet$ (i.e. $dC^n = C^{n,n}$). But the diagonal of $X\_{p,q}$ is just the bar resolution of $G$, so we conclude that the inclusion $dX\_\bullet \hookrightarrow X\_{\bullet\bullet}$ induces a chain homotopy equivalence
$$\operatorname{Tot}C^{\bullet\bullet}\simeq C^\bullet(G,M).$$
Explicit descriptions of the map follow from sufficiently effective proofs of Eilenberg–Zilber. See for instance 8.5.4 in Weibel's *An introduction to homological algebra*. (For your direction, you need to use the Alexander–Whitney map on $\mathbf Z[X\_{\bullet\bullet}]$, because the direction flips when applying $\operatorname{Hom}\_{\mathbf Z[G]}(-,M)$.)


Concretely, the map $\operatorname{Tot}(C^{\bullet\bullet}) \to C^\bullet(G,M)$ takes a $(p,q)$-cocycle $f \colon X\_{p,q} \to M$ to the $(r,r)$-cocycle $X\_{r,r} \to M$ given by
$$\big(n\_0,\ldots,n\_r,h\_0,\ldots,h\_r\big) \mapsto f\big(n\_p,\ldots,n\_r,h\_0,\ldots,h\_q\big),$$
where $r = p+q$. (Here I am working with inhomogeneous cocycles, i.e. $G$-equivariant maps $G^{r+1} \to M$.)


