TITLE: Intuition for Cohomology and Holes in a Space
QUESTION [11 upvotes]: I'm learning that the dimension of the $k^{th}$ De-Rham Cohomology Group (as a vector space over $\mathbb{R}$) of space tells us the number of $k$-dimensional holes in the space. I always found this quite strange, given that the De-Rham Cohomology Group deals with differential forms, a very algebraic object, whereas holes are very geometric. 
So my question is this: Is there an intuitive reason why differential forms on a space and holes in the space have anything to do with each other? Does the presence of holes force the differential form to "dodge" the hole, which in turn changes its properties in ways that we can detect? I know how to formally prove that forms detect holes; that's not what I'm looking for. I am looking for a deeper philosophical answer to the question: what is the intuitive relationship between differential forms and holes? 
Thanks in advance. Appreciate the help!

REPLY [9 votes]: The intuitive reason for why closed forms detect holes is the existence of Stokes' theorem. One of the consequences of Stokes is, for example, that for a closed $1$-form $\omega$ and a path $c$, the value of $\int_{c} \omega$ is invariant if we move $c$ around via an homotopy. If there is another path $c'$ which has a different value for $\int_{c'}\omega$, then we cannot move $c$ around and end in $c'$, thus "there is a hole" of some kind.
It is also arguably the formal reason, since the de Rham map 
\begin{align*}
\Omega_n(X) &\to \mathrm{Hom}(C_n(X),\mathbb{R})\\
\omega &\mapsto (\sigma \mapsto \int_{\sigma}\omega)
\end{align*}
depends on Stokes to even be a chain map. The de Rham theorem then says it induces an isomorphism on the cohomology level, and since the right side is singular cohomology you recover the frequently used "holes" analogy.