homeomorphism between the usual hawaiian ring and the onesided hawaiian ring 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?
 A: 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:

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.
