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Rama Kunapuli (Jul 25 2020 at 21:06):I have a naive question.
Diffeomorphic manifolds have isomorphic de Rahm cohomology groups. De Rahm cohomology groups are also homotopic invariants - that is homotopically equivalent manifolds have isomorphic de Rahm cohomology groups.
Is there a categorical set up where later becomes immediate because of former?
John Baez (Jul 25 2020 at 21:24):The best way I know to show that homotopy equivalent but not necessarily diffeomorphic manifolds have isomorphic de Rham cohomology groups is to show that de Rham cohomology is isomorphic to ordinary cohomology, which is already known to be homotopy invariant.
John Baez (Jul 25 2020 at 21:25):This is called de Rham's theorem, and you need to prove it anyway to understand de Rham cohomology.
John Baez (Jul 25 2020 at 21:26):If you want category theory, you'll get some in the proof of de Rham's theorem: the nice proof uses sheaves!
Rama Kunapuli (Jul 25 2020 at 23:03):First of all thank you for pointing out Categorical aspects of de Rahm's theorem.
SInce, smooth manifolds, cohomology groups categories exist and there are functors that map smooth manifolds and smooth maps to category of real vector spaces and linear maps, I was wondering if there is a categorical way of showing isomorphism of cohomology at least at the underlying topological spaces.
Current method of showing using homotopy operators is nice and direct. But instead of doing all this, if under certain conditions category theory can derive this results from already established proofs from de Rahm theorem. Looks like No, correct?
John Baez (Jul 25 2020 at 23:22):I don't know any "categorical" way of showing that de Rham cohomology is homotopy invariant without showing it's naturally isomorphic to ordinary cohomology, which is homotopy invariant. This fact - that for smooth manifolds, de Rham cohomology is naturally isomorphic to ordinary cohomology - is de Rham's theorem.
Rama Kunapuli (Jul 25 2020 at 23:30):Perfect. Thank you.