Cohomology - futher motivation and definition.

In general, in order to classify somethings, we use an invariant. For example if one is to classify a bunch of books, one can use broad classification such as fiction vs non-fiction. And classification possibilities are more depending on the diversity of the book collection. In case of fiction vs non-fiction classification, the invariant is if a book belongs to fiction or non-fiction. This is nothing new. Similar concepts of classification is used in all fields of sciences including social sciences.

While classifying books - fiction vs non-fiction, we abstract out type of book while removing all other information such as big book, small book, physics book or bible.

In Topology, in a similar fashion we ignore actual geometric shapes, areas etc and focus on whether a space can be deformed continuously into a smaller subset of spaces. When this deformation is taking place, certain things remain invariant.

To see this, assume you have a yet to be inflated balloon where you draw letter "A". Clearly, the letter "A" has one closed loop. As you inflate the balloon, the closed loop becomes larger and gets distorted - but still very much visible. This means such closed loops are invariant under continuous deformation.

Whereas in Topology, the spaces are abstract, when you add extra structure to these spaces as is done in smooth manifolds, the machinery of Linear Algebra can be deployed to perform actual, tangible computations.

For any $k$ form $\omega$, "closed" form means $d\omega=0$. "Exact" form means $\omega=d\tau$ for a form $\tau$ which is a $k-1$ form.

One way to remember these definition is, $d\omega=0$ and since $0$ looks like a closed loop, one can remeber $d\omega=0$ as closed forms.

Let $Z^k(M)$ on a smooth manifold $M$ be vector space of closed forms (think Z for zero) and similarly, let $B^k(M)$ be vector space of exact forms.

Since $d^2\omega = 0$ for any form $\omega$, the exact forms are all closed as $d\omega=d(d(\tau)=d^2\tau=0$.

But not all closed forms are exact forms.

Since, both $B^k(M),Z^k(M)$ are vector spaces, we can form a quotient space $Z^k(M)/B^k(M)$ using the vector space equivalences.

The quotient $H^k(M)=Z^k(M)/B_k(M)$ is called de Rahm Cohomology. This is an invariant of Manifolds under certain conditions which will be noted later.

Since $H^k(M)$ is vector space quotient, for any differential forms $\omega,\omega'$, this means

$\omega' - \omega \in B^k(M)$  means $\omega' \tilde{} \omega$ in $Z^k(M)$.

This means the following relation is satisfied.

$\omega' = \omega + d\nu$.