Let $X = [0,2]$ and $A = \{0,1,2\}$. Prove that $X / A$ is homeomorphic to $C_{1}$ ∪ $C_{-1}$ Let $X = [0,2]$ and $A = \{0,1,2\}$. Prove that $X / A$ is homeomorphic to $C_{1}$ ∪ $C_{-1}$, where $C_1$ and $C_{-1}$ are the circles of radius $1$ centered at $(1, 0)$ and $(-1, 0)$ , respectively.
Ok, here $[0,2]$ has dimension $1$ and $A $ has dimension $0$, since it is a discrete space. Applying the "formula" would $1-0 = 1$ , which coincides with the dimension of the union between the two circunferences. It's right? The formula is the formula of the ligatures, dimensions and quotients.
 A: You can construct an explicit homeomorphism. As a hint, start with explicit homeomorphisms $(0,1)\to C_{-1}\setminus\{\langle 0,0\rangle\}$ and $(1,2)\to C_1\setminus\{\langle 0,0\rangle\}$. You'll use those to construct the homeomorphism you want.
As far as I can tell, all that you've shown is that your two spaces have the same dimension, but that isn't enough to show that they're homeomorphic.

More detail, by request:
First, let me give you a visual idea for why this quotient should "look like" the infinity loop $C_1\cup C_{-1}$. When we take the quotient by $A$, that means that we start treating all the points in $A$ as the same point. Visually, we can imagine starting with the closed interval from $0$ to $2$, then "bend" the interval into an "S" shape so that we can "glue" the ends to the middle. The result is something looking like a "figure 8," and if we twist and stretch it a bit, then it will look just like $C_1\cup C_{-1}$. Some quotients are more difficult to visualize than others, but this one is fairly straightforward.
Before I go on to develop the homeomorphism, take a look at these wonderful illustrations graciously volunteered by Brian M. Scott. He starts by bending the interval until the ends touch, then gluing the joined ends to the middle. This results in the same shape, and will hopefully help you to get a better idea what I'm describing. If you like, you can actually physically do what I'm describing using a piece of string.



Now, if we think about the "bending" process as I described above (with the S curve), or as Brian showed with his pictures, we get an idea how we might map $[0,2]\to C_1\cup C_{-1}$ as a first stage of developing the desired homeomorphism. The points $0,1,2$ will all need to be sent to the origin $\langle 0,0\rangle$ (since that corresponds to the red dot that we labeled the points of $A$ with), one of $[0,1],$ $[1,2]$ will be "wrapped around" $C_1$ and the other will be "wrapped around" $C_{-1}$. Ultimately, it doesn't matter which is wrapped around which, nor which direction we wrap.
The function $f:[0,2]\to C_1\cup C_{-1}$ given by $$f(x)=\begin{cases}\left\langle-1+\cos2\pi x,\sin2\pi x\right\rangle & \text{if }0\leq x\le1\\\left\langle1-\cos2\pi x,\sin2\pi x\right\rangle & \text{if }1<x\le2\end{cases}$$ does the job. It starts at the origin when $x=0$, then traces around $C_{-1}$ counterclockwise until it gets back to the origin when $x=1$, then it traces clockwise around $C_1$ until it gets back to the origin when $x=2$. You should verify that $f$ is continuous and onto, and that it is 1-to-1, except at the points of $A$ (which are all mapped to the origin).
This induces a map $f^*:[0,2]/A\to C_1\cup C_{-1}$, given by $f^*(\overline x)=f(x)$ for any $\overline x\in[0,2]/A$. You should confirm that this is well-defined, 1-to-1, onto, continuous, and an open map. Thus, $f^*$ is the desired homeomorphism.
