General idea
I am trying to understand a bijection between two 'flavours' of the second continuous cohomology group $H^2(\widehat Q; M)$ of the profinite completion $\widehat Q$ of a finitely presented group $Q$ with coefficients in a profinite (but not necessarily finite) $\widehat \Z[[\widehat Q]]$-module $M$.
I am stuck at one specific part.
Setting
Let $\mathcal{P} = \langle a_1, \ldots, a_n\ |\ r_1, \ldots, r_m\rangle$ be a finite presentation for $Q$. This gives a short exact sequence of groups ${1 \to R \to F \to Q \to 1}$ where $F=F(a_1, \ldots, a_n)$ and $R=\langle\langle r_1, \ldots, r_m\rangle\rangle_F$ and a short exact sequence of profinite groups $$1 \to \overline R \to \widehat F \xrightarrow{\pi} \widehat Q \to 1,$$ where $\overline R$ is the closure of $R$ in $\widehat F$.
Also, $\mathcal{P}$ gives a map $d_2: C_2 \to C_1$ of free $\Z Q$-modules $C_2 = \bigoplus_{i=1}^m \Z Q\cdot R_i$ and $C_1 = \bigoplus_{i=1}^n \Z Q\cdot A_i$, coming from the cell complex of the universal cover $C_\mathcal{P} = \widetilde{K_\mathcal{P}}$ of the presentation $2$-complex $K_\mathcal{P}$. After applying functor $$\widehat\Z[[\widehat Q]] \otimes_{\Z[Q]}(-):\ \Z[Q]\text{-modules} \to \widehat\Z[[\widehat Q]]\text{-modules}$$ this gives us a map of free $\widehat\Z[[\widehat Q]]$-modules $\widehat d_2: \widehat C_2 \to \widehat C_1$. If the group $Q$ is residually finite, $\Z[Q]$ embeds into $\widehat\Z[[\widehat Q]]$, so $C_i$ embeds into $\widehat C_i$ and so $\widehat d_2$ is actually an extension of map $d_2$.
Specific problem
I am trying to prove the following statement.
Given a continuous homomorphism of $\widehat Z[[\widehat Q]]$-modules $f: \widehat C_2 \to M$ such that $f(X) = 0$ for all $X \in \ker \widehat d_2$, we can define a continuous homomorphism $g: \overline R \to M$ of profinite groups by setting $g: x r_i x^{-1} \mapsto f(\pi(x)\cdot R_i)$ for $i = 1, \ldots, m$ and $x \in F$ (and extending it continuously to all of $R$).
Why I believe this is true
This is essentially proving that for an $\widehat F$-module $M$ the map $$H^1(\overline R; M)^{\widehat F} \to H^2(\widehat Q; M^{\overline R})$$ coming from the spectral sequence is surjective.