DeRham cohomology

Definition

Let $Ω^{k} (M)$ be the space of smooth $k$ -forms on a smooth manifold $M$ .

A form $ω \in Ω^{k} (M)$ is closed if $d ω = 0$ , where $d$ is the exterior derivative.
A form $ω \in Ω^{k} (M)$ is exact if $ω = d η$ for some $(k - 1)$ -form $η$ .

The De Rham cohomology group is defined as:

H_{dR}^{k} (M) = \frac{\ker (d : Ω^{k} (M) \to Ω^{k + 1} (M))}{im (d : Ω^{k - 1} (M) \to Ω^{k} (M))} .

This means:

$\ker (d)$ : All closed $k$ -forms.
$im (d)$ : All exact $k$ -forms.

The quotient $H_{dR}^{k} (M)$ measures how many closed $k$ -forms are not exact. These equivalence classes are the cohomology classes.

How to find a closed p-form which is not exact?

@baez1994gauge
The trick is to use Stokes’ theorem. Suppose $S \subseteq M$ is a circle embedded in $M$ . If $ω \in Ω^{1} (M)$ equals $d f$ for some function $f$ , then Stokes’ theorem implies

\int_{S} ω = \int_{S} d f = \int_{\partial S} f = 0

because $\partial S$ is the empty set! So if we can find a circle $S \subseteq M$ with

\int_{S} ω \neq 0,

we automatically know that $ω$ is not exact.
This idea can be generalized to $p$ -forms. For examples, we can use spheres to find closed 2-forms which are not exact.

Indeed, the converse is also true: if $\int_{S} ω = 0$ for every compact oriented submanifold $S$ of $M$ then $ω$ is exact. See @baez1994gauge page 125.