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Im trying to understand the proof of Lemma 2.3 of Milnor and Kervaire: Groups of homotopy spheres I.

Suppose we have a simply connected manifold $M$ which bounds a contractible manifold $W'$. Then the Lemma says, that $M$ is h-cobordant to the sphere $S^n$.

The idea is simple and clear. We imbed an (n+1)-disk into $W'$ and remove its interior. Then we get a simply connected manifold $W$ with boundary $M \sqcup S^n$, so it remains to show that both boundary components are deformation retracts of $W$. Then he uses the relative homology sequences of the pairs $(D^{n+1}, S^n)$ and $(W',W)$ and the 5-Lemma to show that the inclusion $i \colon S^n \hookrightarrow W$ induces an isomorphism on the homology.

Since I dont understand where the necessary isomorphisms come from, I thought about another way to prove this and wonder, why it isnt possible:

Since the spaces are good enough, we can use the fact, that \begin{equation*} H_k(W,S^n) \cong \widetilde{H}_k (W/S^n) \end{equation*} But isn't $W/S^n \simeq W'$ since we essentially just collaps the imbedded disk to a point? Then we use the fact that $W'$ is contractible and get \begin{equation*} H_k(W,S^n) \cong \widetilde{H}_k (W/S^n) \cong \widetilde{H}_k (W') \cong 0 \end{equation*} From there it follows by well known arguments, that we have an h-cobordism.

I would be really happy, if you could find my mistake.

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    This looks fine to me, although you do need to be a little careful since quotienting by a contractible subspace is not always a homotopy equivalence. However, since the ball is smoothly embedded, you can invoke the tubular neighborhood theorem to get things to work. – Cheerful Parsnip Aug 16 '13 at 23:44
  • Thanks for the answer. So it is no problem, that I dont really quotient the ball itself but its boundary-sphere? – unknownMathematician Aug 17 '13 at 11:03
  • Got it, I can use excision for a clean proof. Thanks again for the answer. – unknownMathematician Aug 17 '13 at 22:30

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