Seems like a nice exercise , to dualize a theory, or, in this case , to " anti-dualize" , since cohomology is already a dual to homology. We know the dual, we need the " anti-dual". I assume in the finite-dimensional case, it should not be too hard.

A little elaboration on @mathwonk 's explanations.

A cochain complex is a graded collection of modules ##C^{i}## together with a collection of module homomorphisms ##δ^{p}## that map ##C^{p}## into ##C^{p+1}##
such that ##δ^{p+1}δ^{p} = 0##.

Any cochain complex gives rise to cohomology groups ##H^{p}## of ##p##-cocycles modulo ##p##-coboundaries. Differential forms on a manifold together with the exterior derivative form a cochain complex of modules over the real(or complex) numbers and therefore determine cohomology groups in each dimension ##p##. ##H^{p}_{DeRham}## is the group of closed ##p##-forms modulo the exact ##p##-forms.

One might ask what the relation of these cohomology groups is to singular cohomology with real coefficients. A priori they could be different. DeRham's Theorem says that they are the same even though singular cohomology is defined entirely differently. As @mathwonk explained, if one integrates a ##p##-form over a ##p##-chain one gets a homomorphism of the group of smooth singular chains into the real numbers. This interprets the ##p##-form as a smooth singular cochain. Stokes Theorem says that a closed ##p##- form becomes a closed singular ##p##-cocycle and an exact ##p##-form becomes a singular coboundary. So one has a map of cochain complexes, that is: a collection of linear maps ##C^{p}_{DeRham} → C^{p}_{smooth- singular}## that commute with the coboundary operators. This gives a collection of homomorphisms of cohomology groups ##H^{p}_{DeRham}→H^{p}_{smooth- singular}##. DeRham's Theorem says that these homomorphisms are isomorphisms.

The isomorphism between Czech cohomology and DeRham cohomology that I have seen does not use integration. Rather, it uses a double cochain complex which I am happy to explain if someone wants.

- The map from DeRham cohomology is to smooth singular cohomology i.e. where the simplices are required to be smooth. In general singular theory, simplices need only be continuous. One then proves that smooth singular and continuous singular are isomorphic.

BTW: The wedge product of differential forms determines a product on the DeRham cohomology groups. This product gives more information than the cohomology groups themselves. One might ask whether this product corresponds to another product on singular cohomology groups. The answer is yes. It is called the ""cup product".

One of the advantages of cohomology over homology is this cup product since it gives more information than cohomology alone.

I vaguely remember using cup products to distinguish between spaces in ways cohomology alone could not, i.e., the spaces had identical
cohomology groups but their cup product structures were non-isomorphic ( or whatever category applies).