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Let $N_h$ be a closed non-orientable 2-manifold of genus $h\geq 1$. I am trying to compute the first homology groups $H_1(N_h)$.

For do so, it is sufficient compute the cellular homology group $H_1^{CW}(N_h)\cong H_1(N_h)$. Since $N_h$ has one $0$-cell $\{p\}$, $h$ $1$-cells $\{a_1,\ldots, a_h\}$ and one $2$-cell $\{F\}$ the cellular chain complex: $$\ldots \longrightarrow C_3^{CW}(N_h)\stackrel{d_3}{\longrightarrow} C_2^{CW}(N_h) \stackrel{d_2}{\longrightarrow}C_1^{CW}(N_h) \stackrel{d_1}{\longrightarrow}C_0^{CW}(N_h) \stackrel{d_0}{\longrightarrow}0,$$ is equivalent to: $$\ldots \longrightarrow 0\stackrel{d_3}{\longrightarrow} \langle F\rangle_{ab} \stackrel{d_2}{\longrightarrow} \langle a_1,\ldots, a_h\rangle_{ab} \stackrel{d_1}{\longrightarrow}\langle p\rangle_{ab}\stackrel{d_0}{\longrightarrow}0,$$ where $\langle S\rangle_{ab}$ denotes the free abelian group generated by a set $S$.

The map $d_1$ is the trivial map. And for some reason, the map $d_2$ is given by: $$d_2(F)=2(a_1+\ldots+a_h)$$ From here, it is pretty cleat that $\textrm{im}(d_2)$ is isomorphic to $2\cdot\mathbb Z$ and: $$H_1(N_h)\cong\frac{\mathbb Z\oplus \stackrel{(h}{\ldots}\oplus \mathbb Z}{2\cdot\mathbb Z}\cong \mathbb Z \oplus \stackrel{(h-1}{\ldots}\oplus \mathbb Z\oplus \mathbb Z_2.$$

Why $d_2(F)=2(a_1+\ldots+a_h)$? I am not able to get this expression from the definition of the cellular map $d_n$:

$$d_n(\underbrace{e_\alpha^n}_{n-cell})=\sum_{e_\beta\in ``(n-1)-cells``}e\beta\cdot d_{\alpha\beta}$$

where $d_{\alpha\beta}$ is the degree of the map: $$S_\alpha^{n-1}\stackrel{attaching\, map}{\longrightarrow} X^{n-1} \stackrel{quotient\, map}{\longrightarrow} X^{n-1}/X^{ n-2}\stackrel{homeomorphism}{\longrightarrow} \bigvee_{k=1}^hS^{n-1}_k\stackrel{\beta-projection\, map}{\longrightarrow}S^{n-1}_\beta\qquad (\star)$$ Any help would be appreciated.

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  • $\begingroup$ You should probably mention that $N_h$ is a $2$-manifold. Also, the boundary map computation follows because each $1$ cell appears twice on the boundary of the $2$-cell with the same orientation. $\endgroup$ Commented May 14, 2022 at 17:23
  • $\begingroup$ @CheerfulParsnip many thanks for your comment, your are right my question is about 2-manifolds! Anyways, I cannot see why "1 cell appears twice on the boundary of the 2-cell with the same orientation" implies that the map has degree two. How can I derive from the expresion $(\star)$? Is a direct consequence of how the attaching map is defined? $\endgroup$
    – FUUNK1000
    Commented May 15, 2022 at 8:27

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You need to know how the $2$-cell $F$ is attached to the $1$-skeleton $\bigvee_{k=1}^hS_k^1$, namely by the word $a_1^2a_2^2\dotsc a_h^2$ (this word describes the homotopy class of the attaching map in the fundamental group $\pi_1(\bigvee_{k=1}^hS_k^1)=F(a_1,\dotsc,a_k)$, the free group on $a_1,\dotsc,a_k$, where $a_i$ denotes the loop traversing $S_i^1$ once counter-clockwise for each $i=1,\dotsc,h$). If you follow this attaching map by projecting onto the $i$-th $1$-cell $S_i^1$, you end with the map $S^1\rightarrow S_i^1$ represented by the word $a_i^2$, i.e. the map that traverses the circle $S_i^1$ twice in counter-clockwise direction. This map has degree $2$.

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  • $\begingroup$ Thanks for your answer! It is really helpful $\endgroup$
    – FUUNK1000
    Commented May 17, 2022 at 15:40

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