Equivalency of h-cobordisms

Question

I'm reading lectures on the h-cobordism theorem by Milnor and I have a little problem understanding some basic points. I can't understand this theorem , not the theorem , not the proof. I appreciate some help.

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Answer 1

I don't know about Milnor's book, but as an expert on this field and as one of my favorite writers I trust this book to be good. However I heard that in the beginnings of this (rather hard) (...and also exciting) theory people did some mistakes, because it is hard to keep track of all the subtleties, so you should be aware of that. If you want some additional guidance I would also strongly recommend the book which W. Lück is currently writing on this topic.

I will also assume that you know what a cobordism is, and why it is important to talk about $5$-tuples here with fixed inclusions to the boundary!

What Milnor first says in the part you posted, is that two diffeomorphic manifolds are cobordant. That is definitely something you would wish for in such a setting. How does he do it? He defines an $n+1$ dimensional manifold $M\times I$ then he defines a partition of the boundary $M\times 0$ and $M\times 1$ and diffeomorphisms from those boundary parts to $M$ and $M'$ each. He uses the given diffeomorphism for the identification with $M'$. After fixing this data he gets a 5-tuple --- a cobordism.

Now to formulate the theorem in a more modern language maybe, what it says is that the map he defined is a functor $$c: \text{ category of manifolds diffeomorphic to $M$ and morphisms diffeos} \to cob. $$ That means it commutes with diffeomorphisms, i.e. it does not matter if I first make the construction for $h$ then for $h'$ and then compose both resulting cobordisms $c_h,c_{h'}$ as cobordisms or if I immediately construct the cobordism $c_{hh'}$ for the diffeomorphism $hh'$: we will get the same morphism from $M$ to $M''$ both ways (because morphisms in the codomain are equivalence classes of cobordisms). So for that to be true he has to show that both results are equivalent. He does this by taking $c_{hh'}$ on the one side and constructing the manifold $W$ on the other side, which corresponds to the composition of the two cobordisms (which means he glues them together via $h$). With those being equivalent the Theorem now follows.