Minimal definition of h-cobordism

Question

Let $(W,M,N)$ be a cobordism between manifolds $M,N$. If the inclusion $M \to W$ is a homotopy equivalence, is the inclusion $N \to W$ also a homotopy equivalence?

Answer 1

(I am assuming they are orientable).If everything is simply connected then it is true. Observe that Alexander duality tells us that $H^k(W,M)=H_{n-k}(W,N)$. So if $M\to W$ is homotopy equivalence, then $H^k(W,M)=0$. So now if we look at the long exact homology sequence of the pair $(W,N)$ then inclusion map induced isomorphism, and by Hurewicz theorem it is in particular homotopy equivalence.{We use the simply connectedness for the last part for using Hurewicz theorem (Ref:corollary 4.33 Hatcher)}.

Freedman proved that every homology 3 sphere $M$ bounds a contractible four manifold $W'$ (not necessarily smooth). Let $W= W'- \{4-ball\}$ (where $W'$ is smooth). Now observe that $W$ is simplyconnected, since adding a 4 ball doesn't change the fundamental group. One of the boundary componenet is now $N=S^3$. Also observe that $H_i(W,S^3)=H_i(W/S^3)$[by excision] and $H_i(W/S^3)=H_i(W')$. And $W'$ is contactible. So the long exact homology sequence of $(W,S^3)$ with Hurewicz theorem tells us that is homotopy equivalence. But $M\to W'$ is not since $M$ is not simply-connected.