TITLE: Integral classes in de Rham cohomology
QUESTION [23 upvotes]: 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.

REPLY [2 votes]: This has nothing to do with poincare poincare duality or even de rham cohomology.
Let $X$ be any space. By the universal coefficient theorem the canonical map $$H^p(X,\mathbb{Z}) \to \operatorname{Hom}_{\mathsf{Ab}}(H_p(X,\mathbb{Z}),\mathbb{Z})$$
Is surjective. Moreover (also by the universal coefficient theorem if you like) we have
$$H^p(X,\mathbb{R}) \cong H_p(X,\mathbb{R})^*$$
Together we get that $\alpha \in H^p(X,\mathbb{R})$ comes from an integral cohomology class iff the induced linear form $\tilde{\alpha}:H_p(X,\mathbb{R}) \to \mathbb{R}$ is a linear extension of a form $H_p(X,\mathbb{Z}) \to \mathbb{Z}$. In other words:

A real cohomology class is integral iff the corresponding linear form takes integer values when restricted to the lattice $H_p(X,\mathbb{Z})/{\operatorname{Torsion}} \subset H_p(X,\mathbb{R})$