# Intuition of de rham cohomology

I'm studying differential geometry, I find it very interesting however the teacher doesn't give much motivation for new definitions, theorems,...

The last topic of the course dealt with differential forms, Stoke's theorem and de Rham cohomology. I understand that differential forms are able to generalize theorems from calculus. However I don't really see what de Rham cohomology does exactly. We really did see only a few facts without much explanation. I've read this nice answer Intuitive Approach to de Rham Cohomology on a similar question. It well explains the ability of 1-forms to spot holes in a manifold. However it says nothing about higher forms and their uses. Can someone explain the idea of cohomology?

There is also an exercise about this: Given two smooth maps $f,g: S^3 \rightarrow S^2$, do they agree at the level of de Rham cohomology? So are the induced maps $H(S^2) \rightarrow H(S^3)$ the same? From what I know, I would say no because $H^1(S^2),H^1(S^3) \cong \mathbb{R}$ and the linear maps $H(S^2) \rightarrow H(S^3)$ are thus isomorphic to $\mathbb{R}$. So I would say there are many maps which induce different maps at the level of de Rham cohomology. Is this correct? What is the intuition behind this exercise?

• Actually $H^1(\mathbb S^2) = H^1(\mathbb S^3) = 0$, not $\mathbb R$. (So the inducd map on $H^1$'s are all zero).
This is not a good example to consider. $H^1(S^2) = 0 = H^1(S^3)$, $H^2(S^2) \cong \Bbb R$ and $H^2(S^3) = 0$, and, similarly, $H^3(S^2) = 0$ and $H^3(S^3)\cong\Bbb R$. So even though there are an integer's worth of homotopy classes of maps $S^3\to S^2$, they all induce uninteresting (i.e., $0$-) maps $H^*(S^2)\to H^*(S^3)$.
You might learn more from considering different maps $S^1\times S^1\to S^1\times S^1$, where you can explicitly compute the deRham cohomology and the induced maps.