Let $M$ be a connected topological $n$-manifold (not assumed to be compact or boundaryless) and let $D$ an embedded closed $n$-disc. In this situation, there is an inclusion map $S^{n-1} = \partial D \hookrightarrow M\setminus D^{\circ}$.

For which $M$ is the inclusion nullhomotopic?

One obvious example of such a manifold is $S^n$. Are there others?

If the inclusion is nullhomotopic, then $M$ is homotopy equivalent to $M\setminus D^{\circ}$ with an $n$-cell attached by a constant map, i.e. $M$ is homotopy equivalent to $(M\setminus D^{\circ})\vee S^n$. Therefore $H^n(M) \cong H^n((M\setminus D^{\circ})\vee S^n) = H^n(M\setminus D^{\circ})\oplus H^n(S^n) = \mathbb{Z}$ so $M$ is necessarily closed and orientable.

There is also an isomorphism of rings $H^*(M) \cong H^*(M\setminus D^{\circ})\oplus H^*(S^n)$ from which it follows that $H^i(M)$ has no free part for $0 < i < n$ (otherwise Poincaré duality would not hold). By using $\mathbb{Z}_m$ coefficients, the same argument shows that $H^i(M)$ also has no $\mathbb{Z}_m$ part for $0 < i < n$. Therefore $M$ is an integral homology sphere.

Are there any (non-trivial) integral homology spheres with the desired property?


Let $M$ be such an integral homology sphere, with fundamental group $\pi$. As you have said, $M$ is homotopy equivalent to $(M \setminus \mathring{D}) \vee S^n$, so its universal cover $\widetilde{M}$ is homotopy equivalent to $(\widetilde{M} \setminus \pi \mathring{D}) \vee \bigvee^\pi S^n$, and so $H_n(\widetilde{M};\mathbb{Z}) = \mathbb{Z}[\pi]$. But as $\widetilde{M}$ is a connected $n$-manifold its top-dimensional homology must be at most rank 1, so $\pi$ must be trivial.

  • $\begingroup$ Thanks for your answer. I really should have thought to do this. Just a couple of questions about notation. Is $\pi D^{\circ}$ the image of an open disc in $\widetilde{M}$ under the action of the deck transformations (equivalently, the preimage of $D^{\circ}$ under the covering map)? Also, does $\bigvee^{\pi}S^n$ just denote a wedge of $|\pi|$ spheres? I would normally expect to see $\pi$ as a subscript rather than the superscript, that's why I ask. $\endgroup$ – Michael Albanese Oct 2 '17 at 11:28
  • $\begingroup$ Yes, that is what I meant on both counts. $\endgroup$ – Oscar Randal-Williams Oct 2 '17 at 14:16

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