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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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Oh, that's right, sorry. Deleted. –  Plop Sep 21 '11 at 21:42
    
Th degrees of the map in cohomology you mention are not quite right. –  Mariano Suárez-Alvarez Sep 21 '11 at 23:57
    
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 $0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}^*_X \to 0$ –  Lucas Kaufmann Sep 22 '11 at 12:32
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1 Answer

up vote 6 down vote accepted

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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