I saw this definition from Kodaira's Complex Analysis. I am wondering the relationship between it and the singular cohomology group. I think they should be the same, but has no idea about how to prove it.
Some notation and setup: Let $\mathcal{R}$ be an abritrary Riemann surface. Let $\mathfrak{U} = \{U_m\}$ be a locally finite open covering satisfying:
- each $U_j$ is a simply connected region and
- if $U_j \cap U_k \neq\varnothing$, then $U_j \cup U_k \subset W$ for some simply connected region $W$
and then the 1-cochain $c$ is defined as a real-valued function defined on $\{(j, k)| U_j \cap U_k \neq \varnothing\}$ satisfying:
- $c(k, j) = -c(j, k)$
- $c(i, j) + c(j, k) + c(k, i) = 0$ if $U_i \cap U_j \cap U_k \neq \varnothing$
if $c(j, k) = d(j) - d(k)$ for some real-valued function $d: \Bbb{N} \to \Bbb{R}$, then $c$ is said to be cohomologous to $0$(exact). Denote all the 1-cocycles as $L(\mathfrak{U})$ and all exact 1-cocycles as $E(\mathfrak{U})$.
The first cohomology group $H^1(\mathfrak{U})$ of the open covering $\mathfrak{U}$ is defined as $L(\mathfrak{U})/E(\mathfrak{U})$.
Then the author states the de Rham's Theorem:
$H^1_d({\mathcal{R}}) \cong H^1(\mathfrak{U})$
Because normally de Rham's theorem is stated for singular cohomology group, I think the cohomology group $H^{1}(\mathfrak{U})$ defined should be something related to the singular cohomology group and the author choose this definition for simpleness. I want to figure out if this definition is equivalent to the normal definition of singular cohomology group but has no idea.
Any help would be appriciated. Thanks!