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Let $m\geq 1$ and $K\subseteq \mathring{\mathbb{D}^m}$ a convex, compact subset. Here $\mathring{\mathbb{D}^m}$ is the interior of the m-dimensional disk. Let $x\in K$.

I am trying to argue that for $q\geq 0$ we have $$H_q(\mathbb{R}^m,\mathbb{R}^m\setminus K)\cong H_q(\mathring{\mathbb{D}^m}, \mathring{\mathbb{D}^m}\setminus x).$$

By excision it holds that $H_q(\mathbb{R}^m,\mathbb{R}^m\setminus K)\cong H_q(\mathring{\mathbb{D}^m}, \mathring{\mathbb{D}^m}\setminus K)$. Now I am trying to argue that $H_q(\mathring{\mathbb{D}^m}, \mathring{\mathbb{D}^m}\setminus K)\cong H_q(\mathring{\mathbb{D}^m}, \mathring{\mathbb{D}^m}\setminus x)$. Why does that hold? I thought that I could note that $\mathring{\mathbb{D}^m}\setminus x \simeq \mathring{\mathbb{D}^m}\setminus K$ and then use the long exact sequence for a triple. But this answer makes me sceptical that $\mathring{\mathbb{D}^m}\setminus x$ and $\mathring{\mathbb{D}^m}\setminus K$ are homotopy equivalent.

Margaret
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2 Answers2

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For the long exact sequence of a pair strategy, one would specifically need the inclusion $D^\circ_m\setminus K\to D^\circ_m\setminus\{x\}$ to be a homotopy equivalence. Well, one could replace $x$ with any point $x'$ of $K$ via a homeomorphism $D^\circ_m\setminus\{x\}\cong D^\circ_m\setminus\{x'\}$. If $x'$ is chosen to be an interior point of $K$ then one can simply use a radial contraction $D^\circ_m\setminus K\simeq D^\circ_m\setminus\{x'\}$. Because $K$ is compact convex with an interior point, its boundary is a sphere of the correct dimension and there is a continuous retraction $D^\circ_m\setminus\{x'\}\to D^\circ_m\setminus K$ which projects the points of $K$ onto the boundary using the line of sight from $x'$.

I confess that that is handwavy and without clear details. I am tired. On that note, I also have to confess that if $K$ has empty interior it is not obvious to me how to procure a homotopy equivalence here. If we could show that compact convex subsets of $D^\circ_m$ with empty interior are necessarily homeomorphic to disks of a lower dimension, this would be sufficient to conclude by a homology argument similar to the one below but I don't know whether or not that is true. Allegedly this is true. In which case there is a simple homology argument for this given in the final paragraph of this post.

Just for fun, here's how to do all of this without needing to figure out whether or not $D^\circ_m\setminus K\simeq D^\circ_m\setminus\{x\}$. I assume $m\ge1$. Assume first $K$ has nonempty interior.

Embed $\Bbb R^m$ into $S^m$ with stereographic projection. There is an LES: $$\cdots\to H_{q+1}(S^m;\Bbb R^m)\to H_q(\Bbb R^m;\Bbb R^m\setminus K)\to H_q(S^m;\Bbb R^m\setminus K)\to H_q(S^m;\Bbb R^m)\to\cdots$$

Observe that $S^m$ is covered by the open sets (using compactness of $K$) $S^m\setminus\{p\}=\Bbb R^m$ and $S^m\setminus K$, where $p$ is the projection point. Since $K$ is contained in the image of $\Bbb R^m$ (within $S^m$) the boundary is "good" and since $K$ is compact convex its boundary is congruent to $S^{m-1}$ and the generalised Schoenflies theorem kicks in to say $S^m\setminus K$ is a disk.

$H_j(S^m;\Bbb R^m)\cong\widetilde{H_j}(S^m)$ is easy to show. Therefore an application of Mayer-Vietoris shows the map $H_q(S^m;\Bbb R^m\setminus K)\to H_q(S^m;\Bbb R^m)$ is an isomorphism when $q\neq m$ and if $q=m$ then we have $H_m(S^m;\Bbb R^m\setminus K)\cong\Bbb Z^2$ in such a way that the map $H_m(S^m;\Bbb R^m\setminus K)\to H_m(S^m;\Bbb R^m)$ is the projection. Our LES then implies $H_q(\Bbb R^m;\Bbb R^m\setminus K)$ is zero when $q\neq m$, and when $q=m$ $H_q(\Bbb R^m;\Bbb R^m\setminus K)\cong\ker(\Bbb Z^2\twoheadrightarrow\Bbb Z)\cong\Bbb Z$. And it's very easy to show $H_q(D^m;D^m\setminus\{x\})\cong\widetilde{H_{q-1}}(S^{m-1})\cong\widetilde{H_q}(S^m)$ so - we are done, for the case where $K$ has interior.

In the case of empty interior, if we assume $K$ has to be a disk (of lower dimension) then we can conclude as follows; $H_q(\Bbb R^m;\Bbb R^m\setminus K)\cong\widetilde{H_{q-1}}(\Bbb R^m\setminus K)\cong H_q(S^m\setminus K;\Bbb R^m\setminus K)\cong H_q(S^m;\Bbb R^m)\cong\widetilde{H_q}(S^m)$ where the last isomorphisms follow from excision, the long exact sequence of a pair. But again, I don't know if $K$ is necessarily a disk or not, that's just conjecture on my part. It's true for $m=2$ :)

FShrike
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This exhibits why you need the requirement $K\subset\operatorname{int}{(D^m)}$ (i.e. $K$ does not touch the boundary of the disk). Now you can write an explicit homotopy equivalence $\operatorname{int}{(D^m)}\setminus K\to\operatorname{int}{(D^m)}\setminus x$.

Ken
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