Homotopy invariance of de Rham cohomology

Let M,N be smooth manifolds which are homotopy equivalent i.e., there exists smooth maps F:MN and G:NM such that FG is homotopic to identity map on N and GF is homotopic to identity map on M.

Then, Homotopy invariance of deRham cohomology says that the de Rham cohomology groups of M and N are isomorphic.

I am not able to understand the construction given in Lee’s Intoduction to Smooth manifolds.

What is the rough idea behind this proof (or any other proof) of homotopy invariance of de Rham cohomology.

EDIT : Given that M,N are homotopy equivalent as above, we need to prove that HpdR(M) and HpdR(N) are isomorphic. we expect this to come from F:HpdR(N)HpdR(M) and G:HpdR(M)HpdR(N).

I do not understand the idea behind proof of two homotopic maps have induce same deRham cohomology maps.

Once we prove this, then FG and 1N induce same deRham cohomology maps i.e., the composition HpdR(N)FHpdR(M)GHpdR(N) is same as the identity map on HpdR(N) and similarly the composition HpdR(M)GHpdR(N)FHpdR(M) is same as the identity map on HpdR(M). This says that FG=1 and GF=1. Thus, F,G are isomorphisms, inverses to each other, conlcuding that deRham cohomology groups HpdR(M) and HpdR(N) are isomorphic.

How do we prove that two homotopy maps induce same deRham cohomology maps. Let f:MN and g:MN be two homotopy maps, we want to prove that f=g:HpdR(N)HpdR(M) i.e., f(ω)=g(ω)+closed p-form on M when seen as maps Ωp(N)Ωp(M). This means, we are expected to have f(ω)=g(ω)+dη where η is a smooth p1 form.

This gives question of defining a map h:{closed p-forms on N}Ωp(N)Ωp1(M) assigning to each closed p form ω on N a p1 form η on M such that f(ω)=g(ω)+dη.

Then author says it turns out to be far simpler to define h:Ωp(N)Ωp1(M) not with the condition f(ω)=g(ω)+d(hω) for every closed form ω but with a more general condition that f(ω)g(ω)=d(hω)+h(dω)
for every smooth p form. Suppose ω is closed then dω=0 and we get the required condition that f(ω)g(ω)=d(hω).

So, now the question is to define a map h:Ωp(N)Ωp1(M) satisfying the condition as above. How can we think of constructing such map? If we are thinking of going from a p form to a p1 form one obvious thing is to some how integrate this p form. What p form can we integrate here? It is natural to some how integrate the p form f(ω)g(ω) to get a p1 form hω. So, when you reverse the process i.e., when you differentiate you get f(ω)g(ω)=d(hω). This idea is vague and I can not make it any better.

This h is called a homotopy operator in this book.

Any suggestions on how would you think about producing this operator is welcome.


Now I read the book and realize that Lie derivative is introduced after the chapter on cohomology, if the order is reversed there is a very direct interpretation.

You want to prove:

If f0,f1:MN are smooth mapping which are homotopic, then
for all k.

Recall that the induced pullback mapping on Hk is just f0[α]=[f0α] and similar for f1. So you need to show: for any k-form α on N, [f0α]=[f1α], or [f1αf0α]=0.

That is, you want to write f1αf0α as d of something. Note that by the fundamental theorem of calculus,


Here ft is the homotopy between f0 and f1. Of course it is not clear what the right hand side is. We want to give it a more intrinsic interpretation, so that we can check if the right hand side is really d of something.

We let F:M×[0,1]N be the homotopy and ιt:MM×[0,1], ιt(x)=(x,t) be the inclusion. Then ft=Fιt, thus ft=ιtF and

LT is the Lie derivative along the vector T:=t (as a vector field on M×[0,1]). Now the Cartan’s magic formula gives (for any differential form ω, vector fields X)


So we have


Note that the integration are exactly the homotopy operator h constructed: so


So we have the next best thing: the right hand side in general is not d of something, but it is when α is closed. This proves the theorem.

Of course I am just hiding everything in the Cartan’s magic formula. The formula is commonly proved by direct calculation. A more fancy/geometric argument is suggested in Arnold’s classical mechanics here. Note that the latter one also use a homotopy operator.

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