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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.

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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.)

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  • $\begingroup$ I missed this part : "But this is clear: the global sections of these sheaves correspond to locally constant functions" ... Really, thank you. $\endgroup$
    – Analyse300
    Jan 13, 2023 at 17:12
  • $\begingroup$ But we agree on that : it works on every compact complex manifold. $\endgroup$
    – Analyse300
    Jan 13, 2023 at 17:13
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    $\begingroup$ 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? $\endgroup$ Jan 13, 2023 at 18:29
  • $\begingroup$ I don't know. The map $H^{0}(X, \mathbb{C}) \mapsto H^{0}(X, \mathbb{C}^{*})$ is not explicit. $\endgroup$
    – Analyse300
    Jan 14, 2023 at 17:45
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    $\begingroup$ It's perfectly explicit, it is literally just given by exponentiation applied pointwise. $\endgroup$ Jan 14, 2023 at 20:52

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