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I'm trying to find intuition for the homeomorphism $\Bbb R^n \setminus \{\text{pt}\} \approx \Bbb S^{n-1} \times \Bbb R$, but I don't know how this should be true.

I know that $\Bbb S^n \setminus \{\text{pt}\} \approx \Bbb R^n$, but I cannot derive the result I'm looking for from this. The rhs of $$\Bbb R^n \setminus \{\text{pt}\} \approx \Bbb S^{n-1} \times \Bbb R$$ is an infinite annulus with radius $1$ when $n=2$, but I don't see how $\Bbb R^2 \setminus \{\text{pt}\}$ looks like an infinite annulus?

Emil Grönberg
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    Topologically, $\mathbb R$ is the same as $\mathbb R^+$, so replace the $\mathbb R$ in your right-hand side with $\mathbb R^+$, and then you can think of any point in $\mathbb R^n$ minus the origin as corresponding to a point on the unit sphere and a radial distance. – JonathanZ Jul 14 '22 at 16:49
  • What doe you mean by “infinite annulus of radius $1?$” – Thomas Andrews Jul 14 '22 at 16:49
  • Right, it is more obvious that it is homeomorphic to $S^{n-1}\times \mathbb R^+.$ But $\mathbb R^+\cong \mathbb R.$ – Thomas Andrews Jul 14 '22 at 16:51
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    @ThomasAndrews - I think they're thinking about something more like an infinitely long (in both directions) tube, like what paper towels are wrapped around. – JonathanZ Jul 14 '22 at 16:51
  • @ThomasAndrews Wouldn't $\Bbb S^1 \times (0,1)$ be for example the annulus of height one without the top and bottom faces? Using this $\Bbb S^1 \times \Bbb R$ should be an annulus with infinite height in both directions? – Emil Grönberg Jul 14 '22 at 16:51
  • @JonathanZsupportsMonicaC Could you elaborate on what do you mean by "point on the unit sphere and a radial distance"? – Emil Grönberg Jul 14 '22 at 16:52
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    Let $p$ be your point, and $v\neq p.$ Then the map from $\mathbb R^n\setminus {p}\to S^{n-1}\times \mathbb R^+$ is: $$v\mapsto \left(\frac{v-p}{|v-p|},|v-p|\right).$$ This is akin to polar coordinates, in some ways, but instead of an angle, we represent a point as the direction from $p$ (the element of $S^{n-1}$ and the distance. – Thomas Andrews Jul 14 '22 at 16:54
  • @ThomasAndrews - that probably deserves to be a full-on answer. Maybe use $\log(|v-p|)$ so it directly uses $\mathbb R$? – JonathanZ Jul 14 '22 at 16:57
  • @JonathanZsupportsMonicaC OP was looking for intuition, so I assumed an explicit function was not wanted. – Thomas Andrews Jul 14 '22 at 16:58
  • @ThomasAndrews - good point. – JonathanZ Jul 14 '22 at 16:59
  • Technically, $S^1\times(0,1)$ is a cylinder, without boundaries, if we are talking geometric objects - it exists in $\mathbb R^3.$ It is certainly homeomorphic to any annulus, without boundaries, but it is also homeomorphic to a punctured open disk. – Thomas Andrews Jul 14 '22 at 17:08
  • @ThomasAndrews - I grabbed your equation for a full-on answer. Hope you don't mind. – JonathanZ Jul 14 '22 at 17:10
  • One of the remarkable things is that $\mathbb R^n-{0}$ is homeomorphic to $\mathbb R^n\setminus {v\mid |v|\leq 1}.$ There is nothing about the topology of $\mathbb R^n\setminus{0}$ that encodes that we are missing just one point from being a plane. – Thomas Andrews Jul 14 '22 at 17:11
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    @MoisheKohan - Good dup spotting. I'm going to leave my answer up, just for the second paragraph that talks it through, although one good picture is what's really needed (I'm awful at generating math images for computers). If anyone want to grab what I've written and supplement the post you found, or do whatever with it, please feel free. (Though I'm guessing this will just get closed as a dup, and I'll get a few downvotes for crimes agains EoQS. :-) – JonathanZ Jul 14 '22 at 17:17
  • @JonathanZsupportsMonicaC There seems to be a pictorial way to describe the scenario depending on your last comment? – Emil Grönberg Jul 14 '22 at 17:34
  • For $n=2$ consider the usual polar coordinates. – Paul Frost Jul 14 '22 at 17:40
  • @EmilGrönberg - Did you read the second paragraph in my answer? I was referring to that. For $\mathbb R^3$ one could draw a globe, along with a ray coming out of the center of the "Earth", and a little "x" where it intersects the surface of the Earth. BTW, you've been getting a lot of responses, but haven't indicated which, if any, are helping you gain intuition. If you aren't getting answers that are useful to you, you should probably indicate so, and why. – JonathanZ Jul 14 '22 at 17:56
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    For intuition and pictorial explanation in case $n=2$, see https://math.stackexchange.com/questions/269712/finding-a-homeomorphism-mathbbr-times-s1-to-mathbbr2-setminus-0-0?rq=1. Voting to close as a duplicate. – Moishe Kohan Jul 14 '22 at 18:00
  • I think I understood it in a sense. If I describe $\Bbb S^{1} \times (0, \infty)$ as an "infinite cylinder" in $\Bbb R^3$ I can in a sense squish down the cylinder so that it corresponds to concentric circles in $\Bbb R^2$. – Emil Grönberg Jul 14 '22 at 18:00
  • @EmilGrönberg - Me, I'd think of your "infinite cylinder" as being made up of a whole lot of very, very long toothpicks, arranged in a circle. Take the circle at "height" 1 as being the place all the toothpicks are fixed to, we can rotate each of them from pointing up to laying flat in the plane. Maybe imagine a marble with a radius slightly bigger than 1 being dropped from infinity and being pushed down the tube towards our fixed circle at height 1. Of course, toothpicks have finite width, so things get messy inside the circle, but we ignore that. – JonathanZ Jul 14 '22 at 18:11
  • @MoisheKohan I disagree with this being a dupe. The linked duplicate only asks for a technical aspect (what's the inverse of a given homeomorphism?), while this question asks for intuition why the two spaces are homeomorphic. Intuition can be given without an explicit homeomorphism, and an explicit homeomorphism doesn't necessarily help with the intuition. – Vercassivelaunos Jul 14 '22 at 18:38
  • @Vercassivelaunos: Did you see my comments? – Moishe Kohan Jul 14 '22 at 18:45
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    @MoisheKohan I did not. I just looked at the linked dupe. The other one is an actual dupe, though, so everything's fine. – Vercassivelaunos Jul 14 '22 at 19:05

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In order to give a nice, intuitive image, I'm going to replace $\mathbb R$ with $\mathbb R^+$ (they are homeomorphic, via e.g. the $\log()$ function).

So, pick any point in ($\mathbb R^n - \{ p\})$. Then consider the ray starting at $p$ and passing through your point. This determines a point on $\mathbb S^{n-1}$ (the point where the ray passes through the unit sphere centered around $p$), and a value in $\mathbb R^+$, given by the distance from $p$. In $\mathbb R^2$ this would look like the spokes of a wheel, if they were infinitely dense, and also infinitely long.

If it helps you, @ThomasAndrews provided the explicit equations in a comment:

$$v\mapsto \left(\frac{v-p}{|v-p|},|v-p|\right).$$

JonathanZ
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