de Rham cohomology is a formal set-up for the analytic problem: If you have a differential *k*-form on a manifold , is it the exterior derivative
of another differential *k*-form ? Formally, if then . This is more commonly stated as , meaning that if is to be the exterior
derivative of a differential *k*-form,
a necessary condition that must satisfy is that its exterior
derivative is zero.

de Rham cohomology gives a formalism that aims to answer the question, "Are all differential -forms on a manifold with zero exterior derivative the exterior derivatives of -forms?" In particular, the th de Rham cohomology vector space is defined to be the space of all -forms with exterior derivative 0, modulo the space of all boundaries of -forms. This is the trivial vector space iff the answer to our question is yes.

The fundamental result about de Rham cohomology is that it is a topological invariant of the manifold, namely: the th de Rham cohomology vector space of a manifold is canonically isomorphic to the Alexander-Spanier cohomology vector space (also called cohomology with compact support). In the case that is compact, Alexander-Spanier cohomology is exactly singular cohomology.