# Cohomology of compact Riemann surface

Let $$X$$ be a compact Riemann surface. I will denote by $$H^{.}(X, \mathbb{C})$$ (respectively $$H^{.}(X, \mathbb{Z})$$) the Cesh cohomology associated to the constant sheaf $$\mathbb{C}$$ (respectively $$\mathbb{Z}$$). The inclusion of sheaf $$\mathbb{Z} \mapsto \mathbb{C}$$ induces a map $$H^{1}(X, \mathbb{Z}) \mapsto H^{1}(X, \mathbb{C})$$.

Why is this map injective in the case where $$X$$ is a Riemann surface?

I use this to show $$H^{1}(X, \mathbb{Z})$$ is finitely generated over $$\mathbb{Z}$$ and torsion free so I can't use those facts to show the statement above. I try to consider the exponential sequence $$\{0\} \mapsto \mathbb{Z} \mapsto \mathbb{C} \mapsto \mathbb{C}^{*} \mapsto \{0\}$$ but it doesn't help me.

The exponential sequence works fine here. We get a long exact sequence in cohomology

$$0 \to H^0(X, \mathbb{Z}) \to H^0(X, \mathbb{C}) \xrightarrow{\exp(2\pi i \cdot)} H^0(X, \mathbb{C}^{\times}) \xrightarrow{\partial} H^1(X, \mathbb{Z}) \xrightarrow{f} H^1(X, \mathbb{C}) \to \dots$$

and by exactness to show that $$f$$ is injective we need to show that $$\partial = 0$$. By exactness again we need to show that $$\exp(2 \pi i \cdot)$$ is surjective. But this is clear: the global sections of these sheaves correspond to locally constant functions to $$\mathbb{C}$$ and to $$\mathbb{C}^{\times}$$ respectively, so this follows from the fact that $$\exp : \mathbb{C} \to \mathbb{C}^{\times}$$ is surjective.

(Note that we use almost nothing about $$X$$ here; this argument applies to any space with reasonable local connectivity properties (although I'm not sure exactly what is necessary), e.g. any CW complex.)

• I missed this part : "But this is clear: the global sections of these sheaves correspond to locally constant functions" ... Really, thank you. Jan 13, 2023 at 17:12
• But we agree on that : it works on every compact complex manifold. Jan 13, 2023 at 17:13
• You don't need compactness or a complex structure, this argument works on any manifold at least. Maybe it works on any locally connected space? Jan 13, 2023 at 18:29
• I don't know. The map $H^{0}(X, \mathbb{C}) \mapsto H^{0}(X, \mathbb{C}^{*})$ is not explicit. Jan 14, 2023 at 17:45
• It's perfectly explicit, it is literally just given by exponentiation applied pointwise. Jan 14, 2023 at 20:52