Cohomology as measure of connectedness.

Connected spaces:

A Topological space consisting of distinct globs is disconnected. Space that is not disconnected is connected.

There are a few ways to characterize such spaces. Topological space consisting of disjoint union of maximal open sets is disconnected. If only open and closed sets are entire space and null set then such Topological spaces are connected. To see why, if $U,V$ are disjoint open sets whose union is whole Topological space, then clearly $U,V$ are complements of each other, hence they are closed. In such disconnected spaces, more than whole space and null set are both open and closed.

Connectedness is a Topological property as it defined using open sets.

Example: Let $Y=[-1,0)\cup(0,1]$ be topological space. Clearly, $[-1,0),(0,1]$ are open sets in this subspace topology. $Y$  is a separated space as it is an union of disjoint maximal subsets of $Y$.

Computing $H^0(M)$ yields a count of connected components.

First note that there are no exact $0$ forms, since these are $k(=0)-1$ which don't exist.

Hence, $H^0(M)=Z^0(M)$.

Suppose $f$ is a closed zero form on $M$. That is $f$ is a $C*\infty$ function such that $df=0$.

$df = \sum_{i=1}^r  \frac {\partial f}{\partial x}  dx $

$df=0$ implies

$\frac{\partial f}{\partial x_i} = 0$ for each $i$.

which means $f$ is constant in each of the components.

To see this in more detail, let $f$ be a constant function a small region on Manifold $Q=\{q \in Q|f(p)=f(q)\}$. Let $U$ be corresponding chart of $Q$. Since $f$ is constant in $Q$, $df=0$ as each of the partials vanish in this region. Thus $Q$ is open. $f$ continuous means $Q$ is closed. Since $M$ is connected only open and closed sets are whole of $M$ and empty set. Then, $f$ is constant on whole of $M$.

$r$ connected components lead to $H^0(M) \equiv R^r$.