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If $X$ is a compact complex manifold, the exponential sequence gives an injective map $H^1(X,\mathbb{Z}) \to H^1(X,\mathcal{O}_X)$. I think that this shows that $H^1(X,\mathbb{Z})$ is torsion free.

Here is how a thought: Since $H^1(X,\mathcal{O}_X)$ is a vector space, if $n \alpha=0$ on $H^1(X,\mathbb{Z})$ then its image satisfies $n\alpha = 0$ on $H^1(X,\mathcal{O}_X)$ and hence $\alpha = 0$ on $H^1(X,\mathcal{O}_X)$ and consequently on $H^1(X,\mathbb{Z})$.

My question is wheter this argumet is right and if the result is true for every compact orientable manifold (ie, if every such manifold has torsionless $H^1$)

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    Why not Mariano? I'm thinking about the map $H^1(X,\mathbb{Z}) \to H^1(X,\mathcal{O}_X)$ induced by the exact sequence of sheaves $0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}^*_X \to 0$2011-09-22

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The first cohomology of any space has no torsion. One way to see this is via the formula $H^1(X;\mathbb Z)\cong Hom(\pi_1(X),\mathbb Z)$. See my answer to this question.

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Are you sure that $H^1(X,\mathbb Z)\longrightarrow H^1(X,\mathcal O_X)$ is an injective?

Anyway, it is true that $H^1(X,\mathbb Z)$ has trivial torsion by universal coefficient theorem for any manifolds; in fact, $H^1(X,\mathbb Z)=\mathbb Z^{\beta_1}\oplus T_0$, where $\beta_1$ is the first betti number and $T_0$ is the torsion part of $H_0(X,\mathbb Z)$. It is clear that $T_0$ is trivial.