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I am trying to come up with one. I can see that if you reflect the two sided rings and project them onto any one side, you get the one sided ring.

However, I cannot think of a function that does not in a bijective manner. Should we be mapping the odd circles of the two sided to the even ones in the one sided?

Any ideas of a proof that shows that there is a homeomorphism?

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When you say usual Hawaiian ring, do you mean the version that looks like nested figure $8$’s instead of nested circles? – Brian M. Scott Nov 15 '12 at 22:14
yes. Sorry I should have been more specific. Yes, the two sided one looks like the figure 8, while the one sided one looks like nested circles. – user43901 Nov 16 '12 at 0:05

HINT: You’re on the right track. Index the rings in the one-sided Hawaiian earring $0,1,2,\dots$ from the outside in. Now flip the odd-numbered rings over to the other side, and you have the two-sided Hawaiian earring. All you have to do is express this idea as an actual homeomorphism.

Alternatively, show that both are quotients of $S^1\times\Bbb N$ by identifying $\{0\}\times\Bbb N$ to a point. (For the double earring it might be simpler to start with $S^1\times\Bbb Z$.)

Added: To avoid any question, this is the one-sided Hawaiian earring as I understand it:

enter image description here

For each $n\in\Bbb Z^+$ there is a circle of radius $\frac1n$, so the circles shrink down towards the origin. The double earring adds a mirror image copy, mirrored in the $y$-axis.

The circles in the one-sided earring are therefore naturally indexed from the outside in by the positive integers. In the double earring we can keep the same indexing on the righthand side and index the circles on the lefthand side similarly by the negative integers: $-1$ for the circle of radius $1$ centred at $-1$, $-2$ for the circle of radius $frac12$ centred at $-\frac12$, and so on.

I’m suggesting that you construct your homeomorphism to match up the rings like this:

$$\begin{array}{r} \text{One-sided earring}:&1&2&3&4&5&6&7&8&9\\ \text{Double earring}:&1&-1&2&-2&3&-3&4&-4&5 \end{array}$$

I’ve still left some work for you: you still have to deal with the positions of individual points on the circles. I suggest that you map $[0,1)$ continuously and bijectively onto each circle (with $0$ going to the origin) and use that parametrization to identify points on the circles.

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If I flip the odd numbered ones, aren't I making them disappear from the one sided? Also, I am having trouble how to express this as a function. What do you mean by quotient, Brian? I do not think I have heard that term. When you say "outside in", you mean that I label the furthest one at infinity 0, the next one 1, etc. ? – user43901 Nov 16 '12 at 0:43
@user43901: The flipping comment was meant to suggest, not to be a complete recipe; I’ll write a bit more. See this article for the notion of quotient space; you should at least be able to tell from it whether you’ve seen the concept. Outside in: for me the earring as a shrinking nest, so that if you pick one point from each ring, you get a sequence that converges to the common point. Is your version an expanding nest? – Brian M. Scott Nov 16 '12 at 0:52
If it is a homeomorphism, then the inverse also needs to be continuous, right? That is where I am having trouble. I can bijectively define $f: [0, 2\pi) \rightarrow (cos x, sin x)$ with a bit of tinkering on the parametrization of the $cos x, sin x$. But how to do with with $[0,1)$? – user43901 Nov 16 '12 at 2:17
@user43901: These parametrizations aren’t actually homeomorphisms, because $S^1$ isn’t homeomorphic to $[0,1)$; they’re to help you construct the homeomorphism. The only difference between using $[0,1)$ and using $[0,2\pi)$ is a scale factor. I recomment mapping $[0,1)$ to circle $n$ on the right by sending $t=0$ to the origin and mapping continuously and uniformly clockwise around back to the origin as $t\to 1$. Do the same thing on the left side, but wrap counterclockwise, so that your parametrizations are mirror images of the corresponding ones on the right. Then define your ... – Brian M. Scott Nov 16 '12 at 2:24
... homeomorphism by saying what it does to a point $\langle n,t\rangle$, where $n$ is the circle number and $t$ is the parameter in $[0,1)$ identifying the point on its circle. – Brian M. Scott Nov 16 '12 at 2:25

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