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Can we find a homeomorphism from the square $Q_2$ of side length $2$ centered on the origin and the unit circle $S^1$?

We can easily define a map $r:Q \longrightarrow S^1$ by

$$(x,y) \mapsto \bigg(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} \bigg)$$ which is a radial projection onto the unit circle, but how can we define $r^{-1}$ and show that it is continuous?

Thoughts

I think we may define the inverse map as

$$(x,y) \mapsto \bigg(\frac{x}{\sqrt2\max{│x│, │y│}} , \frac{y}{\sqrt2\max {│x│, │y│}}\bigg)$$At least intuitively this maps to a square, and the $\frac{1}{\sqrt2}$ term scales appropriately. However, how can we demonstrate these are continuous maps, thus demonstrating that $Q_2$ and $S^1$ are homeomorphic?

Any help would be appreciated. Regards, MM.

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Is your goal to construct a homeomorphism explicitly, or just to prove that the two things are homeomorphic? If the latter, then there are clearly nicer ways to do it than by trying to write down a closed-form expression for a homeomorphism. After all, a guitar shape is clearly homeomorphic to a circle, but there would be little hope of writing down a homeomorphism in closed form. –  Ben Crowell Jan 29 '12 at 19:51
    
@BenCrowell, the question says "find a homeomorphism", but, even so, doesn't ask for, as you put it, a "closed-form expression": see e.g. my answer. –  msh210 Jan 29 '12 at 20:03
    
@BenCrowell: The aim of my question is to determine how to prove the spaces are homeomorphic, although I am also interested in how we can show that the map $r$ is continuous. I agree that finding an explicit function $r^{-1}$ is unnecessary. Apologies, I should have been more clear. –  Mathmo Jan 29 '12 at 20:22
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I think it's perhaps easier to apply the closed map lemma to your map and observe that it is a closed continuous bijection, that is a homeomorphism. –  JSchlather Jan 29 '12 at 20:40
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You actually don't need to prove that $r^{-1}$ is continuous. Once you show that it exists, you're done because a continuous bijection between compact Hausdorff spaces is a homeomorphism. –  Qiaochu Yuan Jan 30 '12 at 2:02

4 Answers 4

up vote 5 down vote accepted

Let me outline a proof; I shall leave the details of this proof as exercises.

Exercise 1: Prove that the unit circle $S^1$ is homeomorphic to the space obtained from the unit interval $[0,1]$ by identifying the endpoints $0$ and $1$. (Hint: the exponential mapping $\theta\to e^{i\theta}$ can be composed with a linear mapping to give a homeomorphism.)

Let $C$ be the image of a simple closed curve in the plane, that is, let $C$ be the image of a continuous function $f:[0,1]\to \mathbb{R}^2$ such that $f:(0,1)\to\mathbb{R}^2$ is injective and $f(0)=f(1)$.

Exercise 2: Prove that $C$ is homeomorphic to the unit circle $S^1$. (Hint: show that $f:[0,1]\to C$ induces a continuous bijection $\tilde{f}:S^1\to C$ using Exercise 1 and then show that $f$ is an open mapping using the compactness of $S^1$ and the fact that the plane is Hausdorff.)

We now need an elementary lemma of point-set topology:

Exercise 3: Let $X$ be a topological space and let $X=A\cup B$ where $A$ and $B$ are closed subspaces of $X$. Prove that if $g:A\to Y$ and $h:B\to Y$ are continuous functions into a topological space $Y$ such that $g(x)=h(x)$ for all $x\in A\cap B$, then there is a unique continuous function $f:X\to Y$ such that $f(a)=g(a)$ and $f(b)=h(b)$ for all $a\in A$ and $b\in B$.

Finally, we can prove the result of your question:

Exercise 4: Prove that the unit square is the image of a simple closed curve in the plane and conclude that it is homeomorphic to the unit circle. (Hint: you can use Exercise 3 to "glue" continuous mappings together.)

Exercise 5: Give examples of images of simple closed curves in the plane (using Exercise 3) and conclude (using Exercise 2) that they are homeomorphic to the unit circle $S^1$.

I hope this helps!

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Thanks! This is very thorough and clear. –  Mathmo Jan 30 '12 at 10:50

If you choose to go about it in a computational manner, you can prove that both of the maps are continuous by noting that they are compositions of continuous functions.

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I'm not sure why you need algebraic formulas. Define each map $f$ to be radial projection onto the appropriate curve (from the origin) and find continuity by the property "$\forall$ neighborhood $U$ of $f(x)$ $\exists$ a neighborhood of $x$ whose image is contained in $U$", which is easy to check by eye.

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Many thanks for this contribution. Are we able to be more specific about the radial projection 'from the appropriate curve' and how we can deduce continuity from it? –  Mathmo Jan 29 '12 at 20:23
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Radial projection: Your image is the point on the target curve on the same line through the origin as the point in the domain. Continuity: Pick a neighborhood of that image, and a neighborhood of its preimage small enough (say, of small enough angular length around the origin) that its image is contained in that neighborhood. These are very nice spaces, so such a neighborhood exists by inspection. –  msh210 Jan 29 '12 at 20:33
    
Thanks for your help! –  Mathmo Jan 30 '12 at 10:50

It seems off to me. For example, in your second map, the point $(1,0)$, which is on the unit circle, gets mapped to $(\frac1{\sqrt2},0)$, which is not on the square.

If I were you, I would write out the four different maps for each side of the square - I think it simplifies the process, especially since once you remove the max terms, the function is easier to work with.

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