Computing homology maps with non-integer coefficients

Question

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:

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 $\mathbb{Z}$.

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 $d_1=0$.