Homology of complement of a $m$-sphere in $\mathbb R^n$, with $m<n$

Question

Trying to solve the problem in the title, I found this post where a particular case is described and the answer gives a generalization that I think could help me.

Here is my first doubt: How to see that $S^{n-1}\backslash(\{*\}\cup S^{k-1})$ is homotopic equivalent to $S^{n-k-1}\vee S^{n-2}$?

Assuming this result, I define the set $A$ as one of these spheres together with a small open neighborhood of the point shared with the other sphere. And $B$ the same, but with the other sphere and the open neighborhood corresponding to $A$.

This election is intended to make $A$ and $B$ open sets that can be retracted to the spheres. The union is the space I want to solve ($X$) and the intersection is the gluing point.

This way, Mayer-Vietoris says that we have

$\ldots \rightarrow H_n(\{p\}) \overset{}{\rightarrow} H_n(S^{n-k-1})\oplus H_n(S^{n-2}) \overset{}{\rightarrow} H_n(X) \overset{}{\rightarrow} H_{n-1}(\{p\}) \overset{}{\rightarrow} \ldots$,

And since these components are known, I think I could be able to finish the problem.

Does it seem ok to you? Do you have other suggestions/approaches to solve the original problem?

Thank you in advance.

Answer 1

I haven't read the linked question, so I won't suggest alternative approaches, but:

• Yes, you are using Mayer-Vietoris correctly/usefully. Of course when $k$ is very small you might have to deal with some nontrivial maps, but such is life.

• I believe the homotopy equivalence intuitively works like this. For a visual take $n=4, k=2$. Note $S^{n-1}-\{*\}$ is $\Bbb{R}^{n-1}$. Thicken the $(k-1)$-sphere to a $k$-ball with the origin removed, cut that out of the space (so the origin remains). Homotope the whole thing down to the unit $(n-1)$-ball with a hole in a coordinate plane, the shape of a "$k$-dimensional diameter", except the origin. Finally, expand the hole in the $(n-1)-k$ orthogonal dimensions until it meets the boundary $B^{n-1}$. Doing all this, you are left with an $(n-2)$-sphere that has a linear subspace [segment] running through it, and this is homotopic to $S^{n-k-1}\vee S^{n-2}$ by pulling the subspace segment outside and then homotoping the attaching sphere to a point.

• (A special argument may be needed for $n=2$ to deal with non-connectedness; also $n=1$ might be false as stated)

Answer 2

Use the Mayer-Vietoris sequence with $\mathbb{R}^n-S^m$ and a small neighborhood $N\cong S^m\times D^{n-m}$ of $S^m$. We have $X\cup N=\mathbb{R}^n$, $N\simeq S^m$, and $X\cap N\simeq \partial N=S^m\times S^{n-m-1}$. Then the reduced M-V sequence gives isomorphisms $$\tilde{H}_i(X)\oplus\tilde{H}_i(S^m)\cong\tilde{H}_i(S^m\times S^{n-m-1})$$ for all $i$. You can then use either the Kunneth Theorem or induction on $k$ in $S^m\times S^k$ by representing $S^k$ as a union of two hemispheres and using the M-V sequence to derive the Betti numbers $$b_i(S^m\times S^{n-m-1})=\delta_{0i}+\delta_{mi}+\delta_{(n-m-1)i}+\delta_{(n-1)i}$$ Since $b(S^m)=\delta_{mi}$, we find $$b_i(X)=\delta_{0i}+\delta_{(n-m-1)i}+\delta_{(n-1)i}$$ There is no torsion involved here, so this is sufficient.