Twisted duality for a nonorientable surface

The lifted boundary of the two-cell has $a_i$-coefficient $a_1^2\cdots a_{i-1}^2(1+a_i)$, obtained by differentiating the attaching word one letter at a time, or equivalently by grouping its two successive lifted incidences along $a_i$. Under the orientation action in [F1], each preceding square acts as $1$ and $1+a_i$ acts as $1-1=0$. Thus the twisted $d_2:\mathbb{Z}\to {\mathbb{Z}}^g$ is zero.

Each lifted one-cell has endpoint incidence $a_i-1$, which evaluates to $-2$. Hence $d_1:{\mathbb{Z}}^g\to \mathbb{Z}$ is $(x_1,\dots ,x_g)\mapsto -2{\sum }_i{x}_i$. In particular $H_2(N_g;\mathcal{O})=\ker d_2=\mathbb{Z}$; also $H_1(N_g;\mathcal{O})\cong {\mathbb{Z}}^{g-1}$ and $H_0(N_g;\mathcal{O})\cong \mathbb{Z}\mathrm{/}2$. These include $g=1$, where the middle group is zero.

There is a canonical pairing $\mathcal{O}\otimes \mathcal{O}\to \underset{‾}{\mathbb{Z}}$: after choosing either generator $o_x$, send $o_x\otimes o_x$ to $1$ and extend bilinearly. Replacing $o_x$ by $-o_x$ changes both factors, so the map is independent of the choice; the two monodromy signs cancel, so it commutes with transport. Apply [F3] first with the constant system and then with $\mathcal{L}=\mathcal{O}$. The first target is $\mathcal{O}$, and the second is $\mathcal{O}\otimes \mathcal{O}\cong \underset{‾}{\mathbb{Z}}$, giving the two displayed families. Step 2.1 verifies the top group directly. Degrees outside $0,1,2$ vanish, and no AC beyond that already assumed by [F3] is introduced.