I want to compute, for example, the homology of a genus-$g$ orientable surface $M_g$ with $R$ coefficients, where $R$ is any associative, commutative, and unital ring. The construction of the surfaces would be like Hatcher's illustration below:

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As in the illustration, such a surface can be given a CW structure with one $0$-cell, 2$g$ $1$-cells, and one $2$-cell, with the quotients as indicated by the arrows. Hence, we have a cellular chain complex:

$$0 \xrightarrow{d_3} R \xrightarrow{d_2} \oplus_{i = 1}^{2g} R \xrightarrow{d_1} R \xrightarrow{d_0} 0$$

Now if this were with $R = \mathbb{Z}$, then the computation of the chain maps would go as follows: We only have one $0$-cell, so $d_1 = 0$. Then for each $1$-cell, the generator of the $2$-cell is sent once in positive direction of the $1$-cell and once in the negative direction for the other corresponding $1$-cell, and so the degree of this map is $0$. Here, by degree, we mean the multiplying factor of the induced map $\mathbb{Z} \rightarrow \mathbb{Z}$ from $S^1 \rightarrow S^1$. Thus, since the degree is $0$ for all the $1$-cells, $d_2 = 0$. Hence, we obtain that:

$$\tilde{H}_i(M_g; \mathbb{Z}) = \begin{cases} \mathbb{Z} & \text{for } i = 0,2 \\ \mathbb{Z}^{2g} & \text{for } i = 1 \\ 0 & \text{otherwise} \end{cases}$$

But I'm not sure what to do if $R \neq \mathbb{Z}$. We can't make any arguments about where "the generator" is mapped to since $R$ may not be generated by one element, and we can't make any arguments about degrees since that only applies to $\mathbb{Z}$.

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    $\begingroup$ $R$ itself isn't generated by a single element, but the singular chain groups are defined as $C_i = R\{\text{singular i-simplices}\}$, that is the set of singular simplices are taken as the generators of a free $R$-module, so you can define maps between chain groups by defining them on simplices and extending to linear combinations. Similarly you can define maps on the cellular chain groups by defining them on the cells. $\endgroup$
    – William
    Commented Feb 23, 2019 at 3:32
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    $\begingroup$ Did you mean to say "$\mathbb{Z}$ for $i = 2$"? $\endgroup$
    – William
    Commented Feb 23, 2019 at 4:39
  • $\begingroup$ Yes, thank you, I will add that. $\endgroup$ Commented Feb 24, 2019 at 23:43
  • $\begingroup$ What you have now is still not quite right, because if $X$ is any path-connected space then the $0$-th reduced homology $\tilde{H}_0(X;R)$ always vanishes. You should either take the tilde off of $H$ or only say "$\mathbb{Z}$ for $i=2$". $\endgroup$
    – William
    Commented Feb 25, 2019 at 0:49

1 Answer 1


Your expression of the chain complex distills away a bit too much information, namely the cellular chain groups aren't only copies of $R$ but in fact free $R$-modules on sets of cells. In particular if we let $\sigma_0$ be the $0$-cell, $\sigma_2$ be the $2$-cell, and $\{\alpha_1,\beta_1,\dots,\alpha_n,\beta_n\}$ be the $1$-cells, then our cellular chain complex looks like

$$0 \to R\{\sigma_2\} \to \oplus_{i=0}^n R\{\alpha_i, \beta_i\} \to R\{\sigma_0\} \to 0$$

Then your argument in terms of $R=\mathbb{Z}$ can be translated in terms of these formal symbols. For each $\alpha_i$ (or $\beta_i$), the boundary circle of $\sigma_2$ is attached to $\alpha_i$ (or $\beta_i$) twice, with opposite orientations each time. So

$$ d_2(\sigma_2) = \sum_{i=0}^n 0\alpha_i + 0\beta_i = 0 $$ and hence $d_2$ is the $0$ map because $C_2$ is generated as an $R$-module by $\sigma_2$. Similarly $d_1(\alpha_i) = 0 = d_1(\beta_i)$ for each $i$, so since $C_1$ is generated as an $R$-module by the $1$-cells it also follows that $d_1=0$.

  • $\begingroup$ This is a great explanation. Thank you so much. $\endgroup$ Commented Feb 23, 2019 at 5:13

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