Understanding de Rham's theorem.

Question

I've been trying to understand a proof of the deRham's theorem. The statement can be found here: https://en.wikipedia.org/wiki/De_Rham_cohomology#De_Rham's_theorem

I'm not stating the theorem here because my question isn't about the theorem or about a step in a proof for it.

While I understand the existence of the homomorphism between the de Rham cohomology and the singular cohomology, I can hardly follow the standard proof of this theorem. All proofs use the 5 lemma, the Mayer-Vietoris sequence and much more machinery that I'm having a hard time understanding and then the proof is constructed step wise, considering about 4-5 cases and showing that in each case, the homomorphism is indeed an isomorphism.

My question is: Is there a simpler proof for this theorem? Don't we essentially have to show that a certain map is an isomorphism? Why can't we show that the map is well defined, a homomorphism and bijective? I'm wondering if there is a proof that uses just definitions and the Stokes' theorem to prove the de Rham theorem.

Answer 1

Don't we essentially have to show that a certain map is an isomorphism?

Yes. However there are many ways to achieve the same goal.

Why can't we show that the map is well defined, a homomorphism and bijective?

You do understand that certain proofs are difficult not because mathematicians are nasty people trying to make readers sweat? They are difficult for a reason.

So what do you think the reason is? The answer is that the direct approach is a failure (or at least it leads to an equivalent difficult path). Not everything is easy in maths, unfortunately.

Let me give you an analogy: isn't the Riemann Hypothesis essentialy about solving some equation $\zeta(z)=0$? How hard can it be, right? And yet it is still unsolved (at the time of writing this answer).

I'm wondering if there is a proof that uses just definitions and the Stokes' theorem to prove the de Rham theorem.

There might be some simple way to solve the problem that everyone missed. Indeed, such things happened before (e.g. Erdos proof of Bertrand's postulate). If there is such a solution and you happen to know it then feel free to publish it. You may even win some award. :) But AFAIK no known simple solution exists.