If $M$ is a compact Riemann surface, then $H^1(M,\mathbb R)\cong H^1(M,\mathcal O)$ I am wondering why we have the isomorphism stated in the title. Concretely, we have the following exact sequences of sheaves:
$$0\rightarrow\mathbb R\rightarrow\mathcal O\rightarrow\mathcal O/\mathbb R\rightarrow 0$$
Where $\mathbb R$ denotes the constant sheaf over $M$ and the first map is the canonical inclusion. If we pass to the long exact sequence in cohomology, we have that
$$\dots\rightarrow H^0(M,\mathcal O/\mathbb R)\rightarrow H^1(M,\mathbb R)\rightarrow H^1(M,\mathcal O)\rightarrow H^1(M,\mathcal O/\mathbb R)\rightarrow\dots$$
is exact, but somehow, the groups at the extremes should be trivial whenever $M$ is a compact Riemann surface.
I know this has to do with Hodge theory over compact Kähler manifolds, but although I have been searching for a reference of this fact, I haven't found it.
How can we deduce this result from Hodge theory?
Thanks in advance for your answers.

The first three terms of the long exact sequence already form the short exact sequence $0 \rightarrow \mathbb{R} \rightarrow \mathbb{C} \rightarrow \mathbb{C}/\mathbb{R} \rightarrow 0$, since the global holomorphic functions on a compact complex manifold are constant. Therefore, the map $H^0(M,\mathcal{O}/\mathbb{R}) \rightarrow H^1(M,\mathbb{R})$ factors by $0$ and thus the map $H^1(M,\mathbb{R}) \rightarrow H^1(M,\mathcal{O})$ is injective. Hodge decomposition and Serre duality imply that both are vector spaces of the same dimension, therefore the map is an isomorphism.
