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 $S^{n-1} = \partial D \hookrightarrow M\setminus D^{\circ}$.



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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?



Remove a disc from a manifold. When is the resulting sphere nullhomotopic?
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.


