Which spaces are homotopy equivalent in an h-cobordism

cobordismhomotopy-theorymanifolds

All the definitions I've found for h-cobordisms define it as a cobordism $(W,M,N)$ such that the inclusions $i_M:M\rightarrow W$ and $i_N:N\rightarrow W$ are homotopy equivalences. But this doesn't make sense to me since we can't compose these maps with each other.

So maybe the definition means that $i_M$ is a homotopy equivalence with respect to some implied function $i'_M:W\rightarrow M$ (and $i_N$ is, separately, its own homotopy equivalence). My guess is that we would use some retract from $W$ to the image of $M$ and compose that with the inverse of $i_M$; this would mean (identifying $M$ with its image in $W$) that there is a deformation retract from $W$ to $M$ (and another one from $W$ to $N$).

This seems perfectly reasonable, but that doesn't seem to match this question, which describes an h-cobordism as a cobordism where $M$ and $N$ are homotopy equivalent as subsets of $W$. I think that's implied by the definition I described above (reverse the retraction from $W$ to $M$, then retract $W$ to $N$, and the composition gives a homotopy from $M$ to $N$), but it seems like a homotopy from $M$ to $N$ is a weaker condition.

Are they equivalent? If there is a homotopy from $M$ to $N$ within $W$, then one could take the subset of $W$ of points in the image of the homotopy. This should be its own manifold, which would satisfy the first definition.

Am I on the right track with any of this?

Best Answer

I think you're confused about what a "homotopy equivalence" is. A homotopy equivalence is a map $f:X\to Y$ such that there exists a map $g:Y\to X$ such that $fg$ and $gf$ are homotopic to the identity maps. So a homotopy equivalence is just a single map $f$; you don't have to specify a homotopy inverse map $g$ to call $f$ a homotopy equivalence.

So when someone says $i_M:M\to W$ and $i_N:N\to W$ are homotopy equivalences, it means just that: $i_M:M\to W$ is a homotopy equivalence, and also $i_N:N\to W$ is a homotopy equivalence. It doesn't mean they are homotopy inverses of each other.

I would not worry about the question you linked, which asked (rather bizarrely) for an explanation of $h$-cobordism that is not "math based". The description of $h$-cobordism given in the answer was not intended as a precise definition but was just an attempt to convey some intuition. (Indeed, it is not clear to me what it even means for $M$ and $N$ to be "homotopic in $W$". I guess maybe it means there is a homotopy equivalence $f:M\to N$ such that $i_Nf$ is homotopic to $i_M$. To get such an $f$, you can just take $f=gi_M$ where $g$ is a homotopic inverse of $i_N$. This is definitely a weaker condition than $i_M$ and $i_N$ both being homotopy equivalences, since for instance $W$ could be disconnected with $M$ and $N$ in just one of the components.)

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