Integral classes in de Rham cohomology 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.
 A: I think Robin's result is also clear if we look at the definition of the isomorphism between De Rham and (${\cal C}^\infty$) singular cohomologies. (Ok, we are both saying the same thing, but I like to say it this way.)
Remember (see Dupont, Curvature and characteristic classes, Springer LNM 640) the definition of the isomorphism at the cochain level:
$$
I : \Omega^n (M) \longrightarrow C^n (M) \ , 
$$
Here $\Omega^n(M)$ are the degree $n$-forms on $M$, $C^n(M)$ the ${\cal C}^\infty$ singular $n$-cochains. Then, for $\omega \in \Omega^n (M)$, the value of $I(\omega )$ on a ${\cal C}^\infty $ singular $n$-simplex $\sigma$ is defined as
$$
I (w)_{\sigma}  = \int \sigma^*  \omega \ ,
$$
where the integral is taken over the standard $n$-simplex $\Delta^n$.
Hence, for $[I(w)]$ to be an integral class it suffices that all values of $I(w)$ on cycles $\tau = \sum_i \lambda_i\sigma_i$ are integer numbers; that is, if and only if $I(w)_\tau = \sum_i \lambda_i\int \sigma_i^*\omega \in \mathbb{Z}$ for all cycles $\tau$.
A: Here is an explicit procedure based on the isomorphism between the de-Rham and Cech cohomologies 
for smooth manifolds based on R. Bott and L.W. Tu's book: Differential forms in algebraic topology.
The description will be given for a three form but it can be generalized 
along the same lines to forms of any degree. We suppose that the manifold has a finite good cover.
The data needed is the transition functions between the coordinate charts (which will be denoted by: $U_\alpha$, $U_\beta$, etc.) 
and of course, the coordinate expression of the given form on each chart.
Given a three form F, then by the Poincare lemma, on  $U_\alpha,  F = dB_\alpha$, where $B_\alpha$ are two forms on $U_\alpha$.
Thus by the Poincare lemma on $U_\alpha \cap U_\beta$:
$B_\alpha-B_\beta = dA_{\alpha\beta}$ , where $A_{\alpha\beta}$ are one forms on $U_\alpha \cap U_\beta$.
Since on $U_\alpha \cap U_\beta \cap U_\gamma$:
$d(A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha})=0$
Then by the Poincare lemma
$A_{\alpha\beta}+A_{\beta\gamma}+A_{\gamma\alpha} = d\phi_{\alpha\beta\gamma}$
where: $\phi_{\alpha\beta\gamma}$ are zero forms on $U_\alpha \cap U_\beta \cap U_\gamma$.
Again, Since on: $U_\alpha \cap U_\beta \cap U_\gamma \cap U_\delta$:
$d(\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta})=0$
Then:
$\phi_{\alpha\beta\gamma}-\phi_{\beta\gamma\delta}+\phi_{\gamma\delta\alpha}-\phi_{\delta\alpha\beta} = C_{\alpha\beta\gamma\delta}$
where: $C_{\alpha\beta\gamma\delta}$ are constants.
The differential form F is integral iff:
$C_{\alpha\beta\gamma\delta}= 2 \pi n_{\alpha\beta\gamma\delta}$
where $n_{\alpha\beta\gamma\delta}$ are integers on all quadruple intersections.
A: 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 $H_p(X,\mathbb{Z})/{\operatorname{Torsion}} \subset H_p(X,\mathbb{R})$

A: By Poincare duality a $k$-form will be integral iff its integral over all
(smooth) singular $k$-cycles is an integer.
