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It can be shown that $H_{dR}^*(X,\mathbb R)$ and $H^*(X,\mathbb R)$ (singular cohomology) are isomorphic for smooth manifolds. I was told that under that isomorphism closed differential forms whose integral over the generators of the homology of $X$ returns integer values, lets call them $H^*_{dR}(X,\mathbb Z)$, then correspond to $H^*(X, \mathbb Z)$.

But why does $H^*_{dR}(X,\mathbb Z)$ even carry a ring structure?
If $\alpha, \beta\in H^*_{dR}(X,\mathbb Z)\subset H^*_{dR}(X,\mathbb R)$, then why is $\alpha \wedge \beta \in H^*_{dR}(X,\mathbb Z)$, i.e. why is $\int_C \alpha \wedge \beta\in \mathbb Z$ for $C$ a generator of the homology group of appropriate degree?

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  • $\begingroup$ In the side bar I discovered a similar question which the search function unfortunately didn't show to me. Seems like proving this is not as easy as I expected. If anyone has a more direct solution than in the related question (with less algebraic topology), feel free to answer. $\endgroup$ – klirk May 9 '18 at 14:50
  • $\begingroup$ Does this thread answers your question : math.stackexchange.com/questions/29797/… ? $\endgroup$ – Roland May 9 '18 at 18:07
  • $\begingroup$ This is exactly the thread my comment referred to. $\endgroup$ – klirk May 9 '18 at 21:32

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