We have the exponential short exact sequence for compact complex manifolds. Why does the image of $H^1(X,\mathbb{Z})$ span $H^1(X,O_X)$ over $\mathbb{R}$ if $X$ is kahler? The map is injective, I was wondering why the case can't happen: (for example) that the whole lattice is sent to a real line in $H^1(X,O_X)$ with generators sent to $\mathbb{Q}$ linearly independent irrationals?
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Do you know what the cokernel is? – Mariano Suárez-Álvarez May 14 '14 at 02:36
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@MarianoSuárez-Alvarez I think it is a subgroup of $H^1(X,O_X^*)$. Is there more information from this? – May 14 '14 at 02:40