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If $M$ is a differentiable manifold, De Rham's theorem gives for each positive integer $k$ an isomorphism $Rh^k : H^k_{DR}(M,\mathbb R) \to H^k_{singular}(M,\mathbb R)$. On the other hand, we have a canonical map $H^k_{sing}(M,\mathbb Z) \to H^k_{singular}(M,\mathbb R)$ . Allow me to denote (this is not standard) its image by $\tilde {H}^k_{singular }(M,\mathbb Z)$. My question is : how do you recognize if, given a closed differental $k$- form $\omega$ on $M$, its image $Rh^k([\omega]) \in H^k_{singular}(M,\mathbb R) $ is actually in $\tilde {H}^k_{singular}(M,\mathbb Z)$.

I would very much appreciate a concrete answer, ideally backed up by one or more explicit calculations. Thank you for your attention.

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    Kurt: I think your question would be easier to read if you drop those "singular" subscripts. Sure enough everyone knows that $H^k(M)$ without qualifiers stands for ordinary cohomology. Also, why do you need two letters ($Rh^k$) for the isomorphism between De Rham and singular cohomologies? I think Dupont's notation (just $I$, for "integration") is pretty standard and using standard notations helps everyone to understand better what we mean.2010-10-05
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    Robin has answered your question below. As for calculations, you may want to try these yourself in the case of the circle (e.g. by determining explicitly a closed one-form on the circle who integral around the circle is 1, and proving that this one-form is unique up to adding an exact one-form).2010-10-05

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