Finding a quotient map from a circle to a union of a square and two circles

Question

Let the following subsets of $\mathbb{R^2}$:

$C_1=\{(x,y)\in\mathbb{R^2}:(x-2)^2+y^2=1\}\quad C_2=\{(x,y)\in\mathbb{R^2}:(x+2)^2+y^2=1\}\\ C_3=\{(x,y)\in\mathbb{R^2}:x^2+y^2=9\}\quad Q=\{(x,y)\in\mathbb{R^2}: \max{(|x|,|y|)}=1\}\quad$

and the following topological subspace of $\mathbb{R^2}$:

$X_2=C_1 \cup Q \cup C_2$

Find an equivalence relation on $C_3$ whose quotient is homeomorphic to $X_2$. Is it possible to define an equivalence relation on $X_2$ whose quotient is homeomorphic to $C_3$?

My solution is the following. Could someone check if it is correct? Many thanks.

The idea is to consider three arcs partitioning the circle $C_3$, say:

$\Gamma_1=\{(x,y)=(3\cos{\theta},3\sin{\theta}):\theta \in [0, 2\pi/3]\}\\ \Gamma2=\{(x,y)=(3\cos{\theta},3\sin{\theta}):\theta \in [2\pi/3, 4\pi/3]\}\\ \Gamma3=\{(x,y)=(3\cos{\theta},3\sin{\theta}):\theta \in [4\pi/3, 2\pi]\}$

and identify the extremities of each arcs together in order to define quotient maps between $\Gamma_1$ and the circle $C_1$, $\Gamma_2$ and the circle $C_2$, $\Gamma_3$ and the square $Q$. The equivalence relation, say $\mathcal{R}$, is then defined as:

$$[(x,y)]=\left\{\begin{array}{ll}\{(x,y)\}&(x,y)\in\mathring{\Gamma}_1\cup\mathring{\Gamma}_2\cup\mathring{\Gamma}_3\\\{(3,0),(-\frac{3}{2},\frac{3\sqrt{3}}{2})\}&(x,y)\in\partial\,\Gamma_1\\\{(-\frac{3}{2},\frac{3\sqrt{3}}{2}),(-\frac{3}{2},-\frac{3\sqrt{3}}{2}\}&(x,y)\in\partial\,\Gamma_2\\\{(-\frac{3}{2},-\frac{3\sqrt{3}}{2}),(3,0)\}&(x,y)\in\partial\,\Gamma_3 \end{array}\right.$$

The quotient map $f:C_3\to{X_2}$ is defined as:

$$f(x,y)=f(3\cos{\theta},3\sin{\theta})=\left\{\begin{array}{ll}(2+\cos{3\theta},\sin{3\theta})&\theta \in [0, 2\pi/3]\\(-2+\cos{3\theta},\sin{3\theta})&\theta \in [2\pi/3, 4\pi/3]\\(\cos{3\theta},\sin{3\theta})/\max{(|\cos{3\theta}|,|\sin{3\theta}|)}&\theta \in [4\pi/3, 2\pi] \end{array}\right.$$

The function $f$ is a quotient map as it is continuous, surjective and closed. Furthermore, $\forall(x,y)\in C_3,\,\forall (x',y')\in C_3$,$\quad(x,y)\mathcal{R}(x',y') \Leftrightarrow f(x,y)=f(x',y')$. Therefore, there is an homeomorphism from $C3/\mathcal{R}$ to $X_2$.

Regarding the second question, I don't think that is possible. At least I can't find any quotient map that would work.

Answer 1

Here's an alternative equivalence relation. For the record, I thought of this by drawing out a picture, deciding to map $(\pm 3, 0)$ to itself, and then folding the rest of the circle down onto $X_2$.

Let $(x,y) \sim (x',y')$ if and only if $(x,y) = (x', y')$, or $(x,y),(x',y') \in \{(\frac{3\sqrt{3}}{2},\frac{3}{2}), (\frac{3\sqrt{3}}{2},-\frac{3}{2})\}$ or $(x,y),(x',y') \in \{(-\frac{3\sqrt{3}}{2},\frac{3}{2}), (-\frac{3\sqrt{3}}{2},-\frac{3}{2})\}$. (Note that if you think of $C_3$ as $(3\cos t, 3 \sin t)$, then this connects the points where $t = \frac{\pi}{6}, \frac{11\pi}{6}$ and $t = \frac{5\pi}{6}, \frac{7\pi}{6}$.)

The quotient map is admittedly a bit crazy, and there might be a more compact way to write this, but it certainly does the job. Define $f:C_3 \to X_2$ as

$$f(x,y) = f(3\cos t, 3 \sin t) = \left\{ \begin{array}{ll} (2+\cos(6t), \sin(6t)) & t \in [0, \frac{\pi}{6}) \\ (1, \frac{6}{\pi}t-1) & t \in [\frac{\pi}{6}, \frac{\pi}{3}) \\ (3- \frac{6}{\pi}t,1) & t \in [\frac{\pi}{3}, \frac{2\pi}{3}) \\ (-1, 5 - \frac{6}{\pi}t) & t \in [\frac{2\pi}{3}, \frac{5\pi}{6}) \\ (-2+\cos(6t-5\pi), \sin(6t-5\pi)) & t \in [\frac{5\pi}{6}, \frac{7\pi}{6}) \\ (-1, 7 - \frac{6}{\pi}t) & t \in [\frac{7\pi}{6}, \frac{4\pi}{3}) \\ (\frac{6}{\pi}t-9,-1) & t \in [\frac{4\pi}{3}, \frac{5\pi}{3}) \\ (1, \frac{6}{\pi}t-11) & t \in [\frac{5\pi}{3}, \frac{11\pi}{6}) \\ (2+\cos(6t-10\pi), \sin(6t-10\pi)) & t \in [\frac{11\pi}{6}, 2\pi) \\ \end{array}\right.$$

This graph provides a visualization of the function, color-coding which part of $C_3$ maps to which part of $X_2$: https://www.desmos.com/calculator/wt42ifn2dt

You can verify this is quotient map as above, and it is indeed true that if $(x,y) \sim (x',y')$, then $f(x,y) = f(x',y')$. So $C_3/\sim$ is homeomorphic to $X_2$.

My initial intuition was that you were right about the second part, but I've had an idea. Consider the following equivalence relation on $X_2$:

$(x,y) \sim (x',y')$ iff $x,x' \in \{-3,3\}$ or $x=x'$.

This would essentially be like collapsing $X_2$ down to a line segment and connecting the end points. You could then write a function from $X_2$ to $C_3$ based only on the $x$ coordinate of each point. I think this would be a quotient map.