Integration of a cohomology class over a homology class.

Can anyone explain to me what it is said in the following article : http://indico.ictp.it/event/a06114/material/0/0.pdf , page : $3$, by Mr. Aroldo Kaplan :

The paragraph says :

Stokes : $$\int_M d \omega = \int_{ \partial M }\omega$$ implies : $$d \omega = 0 \ \ \Longleftrightarrow \ \ \int_{ \mathrm{boundary} } \omega = 0 \ \ \ \ \ \ \ \ \ \mathrm{and} \ \ \ \ \ \ \ \ \partial M = 0 \ \ \Longleftrightarrow \ \ \int_M \mathrm{exact} = 0$$ so, defining :

$H^k ( X ) = \dfrac{ \{ \omega \in \bigwedge^k \ : \ d \omega = 0 \} }{ \{ \omega \in \bigwedge^{k} \ : \ \omega = d \phi \} } = \dfrac{ \mathrm{closed} }{ \mathrm{exact} }$ , $\ H_k ( X ) = \dfrac{ \{ M \subset X \ : \ \partial M = 0 \} }{ \{ M \subset X \ : \ M = \partial N \} } = \dfrac{ \mathrm{cycles} }{ \mathrm{boundaries} }$

the bilinear (function?): $$H^k ( X ) \times H_k ( X ) \to \mathbb{R}, \ \ \ \ \ \ \ \ \ \ \ \ ( [ \omega ] , [M] ) = \int_M \omega$$ ( = flux of $\omega$ through $M$ ) is well defined.

So, I still do not understand why: $$H^k ( X ) \times H_k ( X ) \to \mathbb{R}, \ \ \ \ \ \ \ \ \ \ \ \ ( [ \omega ] , [M] ) = \int_M \omega$$ is well defined. Can you explain to me that please ?

• What are you unclear about? Integrating a closed form or over a boundary gives $0$, so the integral $\int_M \omega$ depends only on $[M]$ and $[\omega]$. – anomaly Aug 8 '16 at 21:12
• @anomaly : When is the bilinear well defined ? When we don't integrate a closed form as well as when we don't integrate over a boundary ? In this case : $[ \omega ] \not \in H^k ( X )$ and $[ M ] \not \in H_k (X)$. I don't really understand that clearly. :-) – Lina45 Aug 8 '16 at 21:19
• Bah, I meant exact rather than closed above. – anomaly Aug 8 '16 at 21:26
• I don't know what you mean, but it sounds like you don't understand the definition of $H_*$ and $H^*$. – anomaly Aug 8 '16 at 21:27
• A boundary still has a homology class- the zero class. Likewise the zero cohomology class is the equivalence class of all exact forms. You can say e.g. that a representative for a nonzero cohomology class is closed but not exact. – Max Aug 8 '16 at 21:52

let's prove that $([\omega],[M])=([\omega+d\phi],[M])$.
indeed, $([\omega+d\phi],[M])-([\omega],[M])=\int_M d\phi=\int_{\partial M}\phi=0$, because $M$ is a cycle.
secondly, let's prove that $([\omega],[M])=([\omega],[M+\partial N])$.
indeed, $([\omega],[M+\partial N])-([\omega],[M])=\int_{\partial N}\omega=\int_N d\omega=0$, because $\omega$ is closed.