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Prove that $[0,\infty) \times \mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$. I want to do this using the tools of algebraic topology.

Some thoughts I have on the matter is that if we remove a point from $\mathbb{R}^3$ then it becomes homeomorphic to $S^3$ minus two points.. Is $S^3$ minus two points contractible? I don't know. I'd appreciate if you guys gave me a few different approaches to this problem! Thanks!

edit: $\mathbb{R}^3$ minus a point is not contractible, as $\mathbb{R}^3 \cong S^2 \times (0,\infty)$ and $S^2$ is not contractible.

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Use local homology. The $n$-th local homology at a point $x$ in a topological space $X$ is the relative homology group $H_n(X,X-\{x\})$. Every point in $\Bbb R^3$ has third local homology $\cong\Bbb Z$. But in $[0,\infty)\times\Bbb R^2$ the points in $\{0\}\times\Bbb R^2$ have third local homology zero. So these spaces are not homeomorphic.

All this holds for general manifolds with boundary.

Angina Seng
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  • If you could show me an explicit calculation for the local homology of a point in ${0} \times \mathbb{R}^2$ i'd be eternally grateful. –  Jul 19 '19 at 19:58
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    @MathematicalMushroom In this case, both $X$ and $X-{x}$ are contractible, so $H_n(X,X-{x})$ vanishes due to the long exact sequence. – Angina Seng Jul 20 '19 at 03:58
  • So removing a point in ${0} \times \mathbb{R}^2$ from $[0,\infty) \times \mathbb{R}^2$ leaves it a contractible space? Is this obvious to see for some reason? –  Jul 20 '19 at 04:42
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Remove $$\{0\}\times \mathbb {R}^2$$ from $$[0,\infty) \times \mathbb{R}^2$$ and we are left with a simply connected space.

Removing any plane from $$\mathbb{R}^3$$ and the result is not simply connected.

Thus the two spaces are not homeomorphic.

Mohammad Riazi-Kermani
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    Could you explain why removing ${0}\times \mathbb {R}^2$ from $[0,\infty) \times \mathbb{R}^2$ amounts to removing a plane from $\mathbb{R}^3$ under a homeomorphism. – Parthiv Basu Jul 19 '19 at 18:15
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    Because ${0}\times \mathbb{R}^2$ is homeomorphic to a plane in $\mathbb{R}^3$ – Mohammad Riazi-Kermani Jul 19 '19 at 18:22
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    I feel uneasy about this. What's stopping ${0}\times \mathbb {R}^2$ to be embedded in an extremely complicated manner? I don't think this is rigorous. – Parthiv Basu Jul 19 '19 at 18:36
  • ${0} \times \mathbb{R}^2$ is a natural subset of $\mathbb{R}^3$, we don't have to worry about how it is embedded. –  Jul 19 '19 at 19:56
  • I've downvoted this. I'm willing to retract it if @MohammadRiazi-Kermani provides a proof that the image of ${0}\times\mathbb{R}^2$ in $\mathbb{R}^3$ under any potential homeomorphism disconnects $\mathbb{R}^3$. This looks very hard, unless I miss something. – freakish Jul 19 '19 at 20:47
  • @freakish. I have edited my answer. What do you think now? – Mohammad Riazi-Kermani Jul 19 '19 at 21:09
  • @MohammadRiazi-Kermani I'm sorry, but this variant is definitely not true. The standard embedding of $\mathbb{R}^2$ into $\mathbb{R}^3$ disconnects $\mathbb{R}^3$ into two contractible pieces (hence not simply connected). Also it is not hard to embed it in such a way that $\mathbb{R}^3\backslash X$ is homotopy $2$-sphere (and so simply connected), by "squeezing" the standard embedding into a "bounded" plane (simlarly how you squeeze $\mathbb{R}$ to $(-1,1)$). – freakish Jul 19 '19 at 21:13
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    It may be true that $\mathbb{R}^3\backslash X$ is not contractible. But again: this looks hard. Edit: no, the difference can be contractible. You embed $X$ as a half-plane. – freakish Jul 19 '19 at 21:15
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Consider a slightly generalized variant: $X=[0,\infty)\times\mathbb{R}^n$ and $Y=\mathbb{R}^m$ for any $n\geq 0$, $m>0$.

Your idea about removing points was quite close actually. The proof goes like this:

  1. $Y\backslash\{v\}$ is not contractible for any vector $v\in Y$ (the difference is a homotopy sphere)
  2. $X\backslash\{(0,v_1,\ldots, v_n)\}$ is still contractible because it is a star subset of $\mathbb{R}^{n+1}$, with a base point for example at $(1,0,\ldots,0)$

You don't need any hard calculations of homologies/homotopies. The only thing you need to know is that a sphere is not contractible which is a well known result of algebraic topology (although analytical proofs do exist as well).

freakish
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