Let $\Sigma_g$ be the orientable genus $g$ surface. Therefore, we know that the fundamental group of $\Sigma_g$, $\pi_1(\Sigma_g) \cong \langle a_1, \cdots, a_g, b_1, \cdots, b_g \mid [a_1, b_1] \cdots [a_g, b_g] =1\rangle.$
I want to compute the the cohomology group $H^k(\Sigma_g, \mathbb{Z}_w)$ with the local coefficient $\mathbb{ Z}_w,$ where $w$ is a non constant representation of $\pi_1(\Sigma_g) \to Aut(\mathbb{Z})\cong \mathbb{Z}/2.$
I only know that $C^k(\Sigma_g, \mathbb{Z}_w): = Hom_{{\mathbb{Z}[\pi_1(\Sigma_g)]}}(C^k(\tilde \Sigma_g), \mathbb{Z}_w )\cong Hom_{{\mathbb{Z}[\pi_1(\Sigma_g)]}}(C^k(\mathbb{D}^2), \mathbb{Z}_w) .$ therefore we have,
$$C^k(\mathbb{D}^2): \; \; 0 \to \mathbb{Z}[\pi_1(\Sigma_g)] \stackrel{\partial_2}{\to} \mathbb{Z}[\pi_1(\Sigma_g)] \stackrel{\partial_1}{\to} \mathbb{Z}[\pi_1(\Sigma_g)] \to 0 $$ From the cell structure of $\mathbb{D}^2$, I think $\partial_1$ is $(1-a_1, \cdots, 1-a_g, 1-b_1, \cdots,1-b_g).$ I don't know how to find $\partial_2.$
It'll be great if aneone give me a hint to solve the problem.