Show homeomorphism between two quotient topologies

Question

Consider two subspaces of $\Bbb R^2$ in the usual topology:

Square $I^2$ := $\{(x, y) ∈ \Bbb R^2: 0 \leq x, y \leq 1\}$

Annulus $A$:= $\{(x, y) \in \Bbb R^2: 1 \leq x^2 + y^2 \leq 4\}$,

Define the following equivalence relations $∼$ and $≈$ on $I^2$ and $A$ respectively by

$(x, y) ∼ (x, y)\ \forall (x, y) \in I^2$, $(0, y) ∼ (1, y)$ and $(x, 0) ∼ (x, 1)$ if $0 \leq x, y \leq 1$

$(x, y) ≈ (x, y)\ \forall(x, y) \in A,\ (x, y) ≈ (2x, 2y)$ if $x^2 + y^2 = 1$.

I wish to prove that in the respective quotient topologies of $[I^2]_∼$ and $[A]_≈$ are homeomorphic . It is perhaps helpful to view quotient spaces as homeomorphic to the torus; so intuitively they should be homeomorphic. So, we can show show each quotient space is homeomorphic to the torus. But does there exist an explicit homeomorphism for this or something? Is there a proof that need not necessarily involve the torus or by pictures, if possible hmm~

*Just an additional remark my classmate and I were working on an exercise in a new topology course where we encountered the above problem and just realise that he had a similar post at Show two topological spaces are homeomorphic but I will be eligible and happy to offer a bounty hopefully to an answer that is helpful for us, thank you!

Answer 1

Note: throughout, if $p\in I^2$ or $p\in A$, $[p]$ will denote the equivalence class of $p$ in the respective quotient space.

You can first map $I^2$ to $A$ by $$f(x,y) = ((y+1)\cos(2\pi x), (y+1)\sin(2\pi x)).$$ (To visualize this, think of $A \subset \mathbb{C}$ in the obvious way, and think of $f$ as $f(x,y) = (y+1)e^{2\pi i x}$.) As you see, $\{x=0\}$ and $\{x=1\}$ end up in the same place; in fact, $(0,y)$ and $(1,y)$ end up in the same place for all $y$. (You can see this way that if you only identify $\{x=0\}$ and $\{x=1\}$ in $I^2$, you are already homeomorphic to $A$.) And moreover, $\{y=0\}$ and $\{y=1\}$ end up as the inner and outer circles of $A$, with $(x,0)$ and $(x,1)$ at the same angle for all $x$.

So now if $\pi : A \to T_A$ and $\rho:I^2 \to T_I$ are the quotient maps, $\pi \circ f$ factors through $\rho$ because of how the equivalences match up, so that there is $g:T_I \to T_A$ with $\pi \circ f = g \circ \rho$, so really we have a formula for $g$, namely $g([(x,y)]) = \pi(f(x,y))$.

You can check that $g$ is a homeomorphism, given by $$g([(x,y)]) = [((y+1)\cos(2\pi x), (y+1)\sin(2\pi x))].$$

Answer 2

Hint: Since both spaces are clearly compact Hausdorff, it is enough to find a continuous bijection. Parametrise the annulus by polar coordinates, and then do some scaling.

Answer 3

Let's show that the map $g : I^2/_\sim \to A/_{\approx}$ from @csprun's answer is indeed a homeomorphism: $$g([(x,y)]) = [f(x,y)] = [((y+1)\cos(2\pi x), (y+1)\sin(2\pi x))].$$

First note that $g$ is well-defined:

• if $x,y \in (0,1)$ then $(x,y)$ is the only member of $[(x,y)]$
• if $x = 0$, then $(0,y) \sim (1,y)$ but $$f(0,y) = (y+1)(\cos(0),\sin(0)) = (y+1)(1,0) = (y+1)(\cos(2\pi), \sin(2\pi)) = f(1,y)$$ so in particular $[f(0,y)] = [f(1,y)]$.
• if $y = 0$ then $(x,0) \sim (x,1)$ but $$[f(x,0)] = [(\cos(2\pi x), \sin(2\pi x))] = [2(\cos(2\pi x), \sin(2\pi x))] = [f(x,1)]$$

Now we show that $g$ is continuous. As in @csprun's answer, denote the respective quotient maps by $\pi : A \to A/_\approx$ and $\rho : I^2 \to I^2/_\sim$. Recall that quotient topologies are given by $$\{U \subseteq A/_\approx : \pi^{-1}(U) \text{ open in } A\}, \quad \{V \subseteq I^2/_\sim : \rho^{-1}(V) \text{ open in } I^2\}$$

Notice that $g$ satisfies the identity $g \circ \rho = \pi \circ f$. The function $\pi \circ f : I^2 \to A/_\approx$ is continuous as a composition of two continuous functions. We claim that $g$ is continuous. Let $W \subseteq A/_\approx$ be open, we claim that $g^{-1}(W)$ is open in $I^2/\sim$. For that it is sufficient to show that $\rho^{-1}(g^{-1}(W))$ is open in $I^2$, but this is true since $$\rho^{-1}(g^{-1}(W)) = (g \circ \rho)^{-1}(W) = (\pi \circ f)^{-1}(W), \text{ and this is open in } I^2$$

Now onto the inverse of $g$. We will provide an explicit inverse $g^{-1} : A/_\approx \to I^2/_\sim$ and show that it is continuous. First notice that the inverse of $f$ is given by $f^{-1} : A \to I^2$ $$f^{-1}(x,y) = \left(\frac1{2\pi}\omega(x,y), \|(x,y)\| - 1\right)$$ where $\omega : A \to [0,2\pi]$ is the unique angle which the point $(x,y)$ closes with the positive $x$-axis: $$\omega(x,y) = \begin{cases} 0, &\text{ if } y = 0, x > 0\\ \operatorname{arccot}\frac{x}y, &\text{ if } y > 0\\ \pi, &\text{ if } y = 0, x < 0\\ \pi + \operatorname{arccot}\frac{x}y, &\text{ if } y < 0\\ \end{cases}$$ Note (because of $\omega$) that $f^{-1}$ is not continuous on the line segment $\{x > 0, y = 0\} \cap A$, but it is continuous elsewhere on $A$.

Now we claim that $g^{-1}$ is given by $$g^{-1}([(x,y)]) = [f^{-1}(x,y)]$$ Indeed, this is well defined as for $(x,y)$ on the unit circle we have $$[f^{-1}(x,y)] = \left[\left(\frac1{2\pi}\omega(x,y), 0\right)\right] = \left[\left(\frac1{2\pi}\omega(2x,2y), 1\right)\right] = [f^{-1}(2x,2y)]$$ Notice that $g^{-1}$ satisfies $g^{-1} \circ \pi = \rho \circ f^{-1}$ so we have $$(g^{-1} \circ g) \circ \rho = g^{-1} \circ (g \circ \rho) = g^{-1} \circ (\pi \circ f) = (g^{-1} \circ \pi) \circ f = (\rho \circ f^{-1}) \circ f = \rho \circ (f^{-1} \circ f) = \rho$$ so $g^{-1} \circ g = \operatorname{id}_{I^2/_\sim}$. Similarly we show $(g \circ g^{-1}) \circ \pi = \pi$ so $g \circ g^{-1} = \operatorname{id}_{A/_\approx}$. Hence $g^{-1}$ is really the inverse of $g$. It remains to show that $g^{-1}$ is continuous.

First notice that $\rho \circ f^{-1} : A \to I^2/_\sim$ is continuous. We know that $f^{-1}$ is continuous on $A\setminus \{x > 0, y = 0\}$, so $\rho \circ f^{-1}$ is continuous there as well. For $(x,0) \in A$, $x > 0$ let $((x_\lambda, y_\lambda))_{\lambda \in \Lambda}$ be a net in $A$ which converges to $(x,0)$. We can assume $x_\lambda > 0, \forall \lambda \in \Lambda$. We have $$(\rho \circ f^{-1})(x_\lambda, y_\lambda) = \begin{cases} \left[\left(\frac1{2\pi}\operatorname{arccot} \frac{x_\lambda}{y_\lambda}, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda > 0 \\ \left[\left(\frac12 + \frac1{2\pi}\operatorname{arccot} \frac{x_\lambda}{y_\lambda}, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda < 0 \\ \left[\left(0, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda = 0 \\ \end{cases} \to \begin{cases} \left[\left(0, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda > 0 \\ \left[\left(1, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda < 0 \\ \left[\left(0, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda = 0 \\ \end{cases} = \left[\left(0, \|(x,y)\| - 1\right)\right] = (\rho \circ f^{-1})(x,y)$$ We conclude that $\rho \circ f^{-1}$ is continuous at $(x,0)$. Finally, let $W \subseteq I^2/_\sim$ be an open set and we claim that $(g^{-1})^{-1}(W)$ is open in $A/_\approx$. It suffices to verify that $\pi^{-1}((g^{-1})^{-1}(W))$ is open in $A$, which is true: $$\pi^{-1}((g^{-1})^{-1}(W)) = (g^{-1} \circ \pi)^{-1}(W) = (\rho \circ f^{-1})^{-1}(W), \text{ and this is open in } A$$ Therefore $g^{-1}$ is continuous so we conclude that $g$ is a homeomorphism.

Answer 4

Identifying $\mathbb{R}^2$ with $\mathbb{C}$, we have $A = \{ z \in \mathbb{C} \mid 1 \le \lvert z \rvert \le 2 \}$. Define $$q : I^2 \to A, q(x,y) = 2(x+1)e^{2\pi iy} .$$ This is a well-defined continuous map because $\lvert q(x,y) \rvert = 2(x+1) \in [1,2]$. It is a closed map because $I^2$ is compact and $A$ is Hausdorff. Moreover, $q$ is surjective: If $z \in A$, then $z = \lvert z \rvert e^{2\pi it}$ for some $t \in [0,1]$. Hence $(\frac{\lvert z \rvert -1}{2},t) \in I^2$ and $q(\frac{\lvert z \rvert -1}{2},t) = z$.

So $q$ is a continuous closed surjection, hence a quotient map. Obviuosly we have $q(x,y) = q(x',y')$ if and only if $x = x'$ and either $y = y'$ or $\{y, y' \} = \{0, 1 \}$. Note that the latter implies $(x,y) \sim (x',y')$.

Let $p : A \to T = A/\approx$ denote the quotient map. Then $r = p \circ q : I^2 \to T$ is a quotient map. We claim that $r(x,y) = r(x',y') \Leftrightarrow (x,y) \sim (x',y')$.

"$\Leftarrow$" We have $q(x,0) = q(x,1)$, hence $r(x,0) = r(x,1)$, and we have $q(0,y) = e^{2\pi iy}, q(1,y) = 2e^{2\pi iy} = 2q(0,y)$, hence $r(0,y) = r(1,y)$.

"$\Rightarrow$" In case $q(x,y) = q(x',y')$ we are done. So let $q(x,y) \ne q(x',y')$. Hence w.l.o.g. (1) $\lvert q(x,y) \rvert = 1$ and (2) $q(x',y') = 2q(x,y)$. But (2) implies (3) $\lvert q(x',y') \rvert = 2$ and (4) $y = y'$ or $\{y, y' \} = \{0, 1 \}$. From (1) and (3) we conclude $x = 0$ and $x' = 1$. Then (4) shows that $(x,y) \sim (x',y')$.

Our above claim proves that $r$ induces a homeomorphism $\hat{r} : I^2 /\sim \phantom{.} \to A/\approx$. See also When is a space homeomorphic to a quotient space?.