TITLE: Parabolic cohomology of modular groups and cup-products
QUESTION [7 upvotes]: I am stuck with a technical question concerning parabolic cohomology of modular groups and cup-products on them. Basically, I am trying to understand the appendix about cohomology of Hida's book "elementary theory of $L$-functions and Eisenstein Series" and related ideas, and though Hida's exposition is very detailed, I am not illuminated.
So, $\Gamma$ is a congruence subgroup of $SL_2(\mathbb Z)$ which acts freely on the Poincaré upper half-plane, for instance $\Gamma_1(\nu)$ for $\nu \geq 3$. We choose for each cusp of $\Gamma$ a generator of the subgroup fixing that cusp, and we define a subset $P$ of $\Gamma$ as the set of all conjugate of those elements. We let $M$ (and later $N$) be left $\Gamma$-modules.
So we have the cohomology groups $H^i(\Gamma,M)$ defined as the homology of the usual complex $C^0(\Gamma,M) \rightarrow C^1(\Gamma,M) \rightarrow C^2(\Gamma,M) \rightarrow \dots$ These groups have a geometric interpretation, namely there are canonical isomorphisms $H^i(\Gamma,M) \simeq H^i(Y_\Gamma,\tilde M)$ where $Y_\Gamma$ is the quotient of the upper half-plane by $\Gamma$ and $\tilde M$ the locally free sheaf on it defined by $M$, and the RHS $H^i$ is sheaf cohomology.
To define parabolic cohomology we consider a subcomplex $C^\bullet_P(\Gamma,M)$ of $C^\bullet(\Gamma,M)$ as follows: $C^i_P(\Gamma,M) = C^i(\Gamma,M)$ for $i=0,2,3,4,\dots$ and for $i=1$, we set $$C^1_P(\Gamma,M) = \{u \in C^1(\Gamma,M), u(\pi) \in (\pi-1) M \ \ \forall \pi \in P\}.$$
(To check that $C^\bullet_P$ is a subcomplex we just need to check that the differential sends $C^0$ into $C^1_P$ but this is trivial.) We define $H^i_P(\Gamma,M)$ as the homology of that complex. It is trivial that $H^i_P(\Gamma,M)=H^i(\Gamma,M)$ for all $i \neq 1,2$. Hida also gives a geometric interpretation of the new cohomology groups: there are canonical isomorphisms
$$H^1_P(\Gamma,M) = H^1_! (Y_\Gamma, \tilde M) := Im \left(H^1_c(Y_\Gamma,\tilde M) \rightarrow H^1(Y_\Gamma,\tilde M)\right),$$
$$H^2_P(\Gamma,M) =H^2_c(Y_\Gamma,\tilde M),$$ where $H^i_c$ is the sheaf cohomology with compact support. (The second isomorphism is given in the last assertion of Prop. 2 of the appendix of Hida's book, and the first is Prop. 1 of his paper Congruences of cusp forms and special values of $L$-functions, invent. math 63 (1981).) 
So far, so good. Now the cup-product 
$H^1_c(Y_\Gamma,\tilde M) \otimes H^1_c(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma,\tilde M \otimes \tilde N)$ induces a cup-product $H^1_!(Y_\Gamma,\tilde M) \otimes H^1_!(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma, \tilde M \otimes \tilde N)$ (as explained in Hida's paper, cf (2.2) and the first page of section 2), hence a cup-product
$$H^1_P(\Gamma,M) \otimes H^1_P(\Gamma,N) \rightarrow H^2_P(\Gamma,M \otimes N).$$
Now my question:

How can this cup-product be described in terms of the definition of $H^\bullet_P$ as the homology of $C^\bullet_P$?

That is, how can this cup-product be described in purely group cohomological terms?
In theory, one should be able to answer this question by translating the definition of the cup-product in sheaf cohomology through all the isomorphisms involved, but this looks not very engaging. What worries me more is that I can't seem to be able to guess what the result could be, that is I am not able to define a cup-product $H^1_P \otimes H^1_P \rightarrow H^2_P$ in the only natural way I can think of, that is by defining it at the level of cochains. Indeed, the natural map $C^0(\Gamma,M) \otimes C^1_P(\Gamma,N) \rightarrow C^1(\Gamma,M \otimes N)$ does not seem to have image in $C^1_P$. So, I would be happy with any "natural" definition of the cup-product in terms of cochains, and quite ready to believe it corresponds to the one defined using geometric cup-product.

REPLY [2 votes]: Since I seem to my stuck on something myself, let me take a crack at your problem. This  is just a proposal,  I haven't checked  if this really works.
Let me start with the suggestion in Dan's answer and identify $H^i_P(\Gamma, M)= H^i(X, j_*M)$, here $X=X_\Gamma$ is a smooth compactification of $Y=Y_\Gamma$. I'm conflacting $M$ with $\tilde M$ and dropping various decorations for ease of writing. Then from the triangle
$$j_*M\to \mathbb{R}j_*M\to R^1j_*M[-1]\to $$
we get
$$H^1_P(\Gamma, M)\to H^1(Y, M)\to \bigoplus_p H^1(C_p,M)\to H_P^2(\Gamma,M)\to0$$
where $p$ (resp. $C_p$) are (resp. circles around) the cusps. This can be translated back to group cohomology
$$H^1_P(\Gamma, M)\to H^1(\Gamma, M)\to \bigoplus_p H^1(P_p,M)\stackrel{\partial}{\to}H_P^2(\Gamma,M)\to 0$$
where now $P_p=\pi_1(C_p)$. Given classes $\alpha\in H^1_P(M),\beta\in H_P^1(N)$, their product, whatever it is, would have to be $\partial(?)$. Here is a guess as to what it might be.
Let us assume that the above sequence is induced by an exact sequence of suitable complexes
$$0\to C_P^\bullet(M)\to C^\bullet(M)\to \bigoplus C^\bullet(P_p, M)\to 0$$
(Perhaps the ones  specified in your question will work for this,  I haven't checked.)  Let $\alpha\in C^1_P(M),\beta\in C_P^1(N)$ denote cocycles representing classes above. Then $\alpha,\beta$ can be mapped to cocycles, and in fact coboundaries, $\alpha',\beta'$ in $\oplus C(P_p,-)$. Write $\alpha'=d \alpha''$. Try $\partial(\alpha''\cup\beta')$.