Constructing a Hoemeomorphism between two figure of eights If we have the identification space of $I=[0,1]$ where we say that $x\sim y$ iff $x=y$ or both x and y are in $\{0,\frac{1}{2}, 1\}$. I am trying to create a homemorphism between this and the space $C_1\cup C_2\subset \mathbb{R}^2$ where $C_1$ is the cricle of radius 1 centred at $(1,0)$ and $C_2$ is the circle of radius 1 centred at $(-1,0)$.
Am I right in thinking that both of these spaces are "figures of eight"? so does the following provide a homemorphism between the two?
$f(x)=\begin{cases} \exp(4\pi i x)-1 & x\in[0,\frac{1}{2}) \\ 0 & x=\frac{1}{2} \\ \exp(4\pi i x+\pi i)+1 & x\in(\frac{1}{2},1]\end{cases}$
Cheers
 A: We have $I$ and the relation $x \sim y \Leftrightarrow \begin{cases} x = y \\ x,y \in \{ 0 ,1/2 ,1\} \end{cases}$. Let's call $I/\sim = A$.
Now we use your function $f(x) = \begin{cases} exp[4\pi i x] \quad x \in [0,1/2] \\ exp[4\pi i x + ix] \quad \in [1/2,1]\end{cases}$ and we have to view if $f$ is continuous in $A$, open map and $f^{-1}$ is continuous. So $f$ is a homeomorphism. It's the definition :)
Let's start with the open map :
$I \setminus \{1/2\}$, for the exponential proprieties, is continuous and every open in $A$ is in the form $(a,b)$ with $\begin{cases} a < b < 1/2 \quad \text{ for } 0<a<1/2 \\  a < b < 1 \quad \text{ for } 1/2 < a < 1 \\ \end{cases}$ and $f$ maps they in a not-closed arch and viceversa.
Now, for $1/2$ we see that in the $eight$ we can obtain opens which have inverse image like $(a,b)$ with $a< 1/2 < b$ , $0<b<1/2, 1/2 <a<1$ and $ (0,b<1/2) \cup (a<1/2 , 1/2)$ and other similar.
In the quotient topology they are all open. So $f$ is an open map.
Continuous :
$f$ is continuous in $A$ because $\lim_{x\rightarrow 0^+} f(x) = \lim_{x\rightarrow 1^-} f(x) = \lim_{x\rightarrow (1/2)^+} f(x) = \lim_{x\rightarrow (1/2)^-} f(x) = 0$. In the rest of $A$ is a propriety of the exponential.
Bijection :
The exponential is a bijection and so it's $f$ in $I \{0,1/2,1\}$. But we have $f(0) = f(1/2) = f(1)$. Our $\sim$ make $f$ a bijection because $0 \sim 1/2 \sim 1$ so they are "like the same point" from a quotient view. So even the $f^{-1}$ map, will be a bijection, continuous.
Because $f : A \rightarrow \mathbb{C}$ is an open map, has an inverse open map, bijective, $f$ is a homeomorphism from $A$ to $\mathbb{C}$.
Yes, your $f$ is the homeomorphism! And it makes a $\infty$
I think that you can imagine the image of $f$ like an eight and for simple comprehension that $A$ is an eight. But they are different in the form. $A$ is a quotient space, $f(A)$ is a curve in $\mathbb{C}$, they have also different topology. 
