# Finding a homeomorphism $\mathbb{R} \times S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$

Are there any specific 'tricks' or 'techniques' in finding homeomorphisms between topological or metric spaces?

I'm trying to construct a homeomorphism between $\mathbb{R} \times S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$ but I'm having trouble even visualizing how this is going to work. I'm thinking some sort of stereographic projection from the circle and then including the $\mathbb{R}$ somehow but I'm stuck. Any ideas?

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@Elias: There is a huge difference between $\mathbb R^2\setminus\{(0,0)\}$ and $\mathbb R^2/\{(0,0)\}$. –  Asaf Karagila Jan 3 '13 at 14:48

## 4 Answers

Think about how the plane minus a point looks: it is a disjoint union of concentric circles, each circle being of some radius $r \in (0,\infty)$. (This is how polar coordiantes work.)

This essentially is the same as saying that $\mathbb R^2 \setminus \{(0,0)\} \cong S^1 \times (0,\infty).$ Now you just have to remember that $(0,\infty) \cong \mathbb R$.

Rather than asking about general methods of finding homeomorphisms (which is a very difficult problem when stated in full generally), based on what you have written, I think that perhaps you should practice visualizing different examples of products. E.g. you might want to convince yourself that a torus (the surface of a donut) is homeomorphic to $S^1 \times S^1$, by thinking of it as a circle of cirles (just as $\mathbb R^2\setminus \{(0,0)\}$ is a circle of intervals). You could also consider the complement of a ball of radius $1$ in the ball of radius $2$, and convince yourself that it is homeomorphic to the product of $S^2$ and a closed interval.

One important "background" fact to keep in mind is that all open intervals are homeomorphic, and that all closed intervals are homeomorphic, so that the particular endpoints don't matter. (In different concrete situations, it can be helpful to think in terms of different endpoints. E.g. $\mathbb R = (-\infty,\infty)$, but in my explanation of your problem above, it was more convenient visually to think in terms of the open interval $(0,\infty)$.)

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I don't know any "tricks" for this. Sometimes, as in this case, you can gain some "inspiration" from the composition factors of the domain. Here we have unbounded and periodic factors, which can be regarded as the modulus and argument components of a nonzero point in the complex plane. Based on this, we can try the function $f(x,\theta)=(e^x \cos\,\theta,e^x\sin\,\theta)$, which turns out to do the trick.

However, most of the time you just have to feel your way forward. Look at different ways to decompose spaces into factors, unions, wedges, etc.

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Here’s a more or less pictorial intuition.

For $0\le\theta<2\pi$ let $R_\theta$ be the open ray $\big\{\langle r,\theta\rangle:r\ge 0\big\}$, where $r$ and $\theta$ are polar coordinates. Let $p_\theta$ be the point $\langle 1,\theta\rangle\in R_\theta\cap S^1$. Pivot $R_\theta$ about the point $p_\theta$ until it’s perpendicular to the $xy$-plane, parallel to the $z$-axis. Do this simultaneously to all of the $R_\theta$, and you end up with an open half-infinite cylinder,

\begin{align*} S^1\times(-1,\to)&=\big\{\langle 1,\theta,z\rangle:z>-1\big\}&&\text{(cylindrical coords.)}\\ &=\big\{\langle x,y,z\rangle:x^2+y^2=1\text{ and }z>-1\big\}&&\text{(rectangular coords.)}\;. \end{align*}

This construction leaves the $\theta$ coordinate alone and converts the $r$ coordinate into a $z$-coordinate: $z=r-1$.

Now just stretch the part of the cylinder below the $xy$-plane. There are lots of homeomorphisms that will do this; all you’re doing at this point is finding a homeomorphism between $(-1,\to)$ and $\Bbb R$.

I have no useful answer to the general question; having seen a variety of examples helps, of course. In this case the two keys for me are recognizing that removing one point and removing a closed disk leave homeomorphic spaces, and recognizing that this space that’s left is homeomorphic to an open open annulus, since it’s pretty clear that an open annulus is homeomorphic to $(0,1)\times S^1$, which in turn is homeomorphic to $\Bbb R\times S^1$.

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In general, it is very difficult to produce a specific homeomorphism. Stereographic projection typically refers to the projection of a sphere onto the plane, and the idea won't work quite so well in this context because the cylinder $\mathbb{R} \times S^1$ extends infinitely in two directions. But if we could shrink one end of the cylinder, then will be able to apply a projection mapping.

Consider $\mathbb{R} \times S^1$ as the set of points in $\mathbb{R}^3$ given by: $\{ (x, y, z) \;|\; y^2 + z^2 = 1 \}$. This in an infinite cylinder parallel to the $x$-axis. Step 1. Let $\phi : \mathbb{R}^3 \to \mathbb{R}^3$ be the map that is the identity when $x < 0$ and sends $(x,y,z) \mapsto (\arctan x, y, z)$ when $x \geq 0$. $\phi$ is a homeomorphism $\mathbb{R} \times S^1 \to (-\infty, \pi/2) \times S^1$. Step 2. Let $P$ be the point $(\pi/2, 0, 0)$, and let $\psi : (-\infty, \pi/2) \times S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$ be stereographic projection from $P$. That is, $\psi(x,y,z)$ is the point of intersection of the $yz$-plane and the line joining $P$ and $(x,y,z)$. Thus, $\psi\phi$ is a homeomorphism $\mathbb{R} \times S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$.

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