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I am asked to prove that the cylinder and the punctured plane are homeomorphic.

I understand that I need to find a function that maps every point in the plane to a point on the cylinder.

I can represent every point on the plane in polar coordinates and now I just need to find a mapping from each point in the plane in R^2 to the cylinder in R^3.

I have found that a solution is $f(r\cos\theta,r\sin\theta)=(\cos\theta,\sin\theta,\log(r))$ which makes sense except I cannot understand why $f(r\cos\theta,r\sin\theta)=(\cos\theta,\sin\theta,r)$ would not be a solution. Since $r \in (-\infty,\infty)$, wouldn't this still result in a cylinder?

user7090
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1 Answers1

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Once you figure out how to map a single "ray from the origin" on the plane to its corresponding "line" on the cylinder, you can extend this to the entire plane by putting $e^{i\theta}$ in front of everything (in other words, repeating that map for each angle, and using it to send each ray to its line).

Here is a gif I made (originally from this post) that may help illustrate the map we should think of (I assume that in your problem, we are supposed to consider an infinite cylinder and the entire punctured plane, not a finite cylinder and an annulus, though in fact they are all homeomorphic).

enter image description here

To be specific, let's say that the cylinder we're looking at is the infinite vertical cylinder of radius $1$: $$C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=1\}.$$ The punctured plane is $$P=\{(x,y)\in\mathbb{R}^2:x\neq 0\,\text{ or }\,y\neq 0\}.$$ So, let's figure out how to do one ray and one line. Hopefully you agree that the ray $$r_0=\{(x,0)\in\mathbb{R}^2:x>0\}$$ should be sent to the line $$\ell_0=\{(1,0,z)\in\mathbb{R}^3:z\in\mathbb{R}\}$$ (roughly speaking, because they are both at an angle of $0^\circ$, when viewed from above).

We need to find a homeomorphism from $r_0$ (which is itself homeomorphic to $\mathbb{R}_{>0}$) to $\ell_0$ (which is itself homeomorphic to $\mathbb{R}$). So we just need to find a homeomorphism from $\mathbb{R}_{>0}$ to $\mathbb{R}$.

There are many examples, but one good one is the logarithm. Just like how $0$ is the "neutral element" for addition in the real numbers, $1$ is the "neutral element" in multiplication. And in fact, the logarithm is precisely the map $\log:\mathbb{R}_+\to\mathbb{R}$ that converts multiplication into addition: $$\log(a\cdot b)=\log(a)+\log(b).$$ You can see that $\log(1)=0$, and that $\log(1),\log(2),\log(4),\log(8),\ldots$ are evenly spaced "additively" in $\mathbb{R}$, just like how $1,2,4,8,\ldots$ are evenly spaced "multiplicatively" in $\mathbb{R}_{>0}$.

enter image description here

So, we can surmise that in the homeomorphism $f:P\to C$ that we're thinking of, we are going to have $f(r_0)=\ell_0$, and that the formula for how $f$ acts is $$f(x,0)=(1,0,\log(x)),\;\text{ for all }x>0$$ Can you now modify this to see the formula for the entire map $f$?

(Again, I should say that of course there are infinitely many homeomorphisms from $P$ to $C$, this is just one that's easy to think of / write down.)

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Zev Chonoles
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  • Thank you! I understand why log works. I needed to find a function that takes only positive values(namely r) and maps to $(-\infty, \infty)$, and $\log$ does this nicely. I overlooked the easy fact that $r >0$. – user7090 Feb 09 '14 at 20:02