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Let $D^2 =\{\sqrt{x^2+y^2} \leq 1\} \subseteq \mathbb{R}^2$, i.e. let $D^2$ be the closed disk of radius $1$ in $\mathbb{R}^2$. I want to show that the quotient space $\mathbb{R}^2/D^2$ is homeomorphic to $\mathbb{R}^2$.

I'm having a hard time figuring out how to show this explicitly.

If I'm understanding correctly, I could also say that we are trying to show that $\mathbb{R}^2/\sim $ is homeomorphic to $\mathbb{R}^2$, where the equivalence relation $\sim$ is given by:

$(x_1,y_1) \sim (x_2,y_2) \iff$ either $[(x_1,y_1) \in D^2$ and $(x_2,y_2) \in D^2$] or $(x_1,y_1) = (x_2,y_2)$.

I guess I want to find a continuous map from $\mathbb{R^2} \to \mathbb{R^2}$ that preserves equivalence classes, but I'm not really sure how to proceed.

Any help is appreciated.

TuringTester69
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  • What does $\mathbb{R}^2 / D^2$ mean? Did you perhaps mean $\mathbb{R}^2 \setminus D^2$? – copper.hat Oct 22 '20 at 20:30
  • @copper.hat I am referring to the quotient space – TuringTester69 Oct 22 '20 at 20:32
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    I think he means the quotient set induced by the relation he wrote – Vajra Oct 22 '20 at 20:33
  • Try scaling along straight lines through origin. The disk maps to $0$ and any point outside the disk has a straight line passing through the origin; scale it toward the origin based on how far from the disk the point is (so closer to the disk will be mapped closer to the origin). – Clayton Oct 22 '20 at 20:37
  • Map $D^2$ to $0$ and $\mathbb{R}^2 \setminus D^2$ to $\mathbb{R}^2 \setminus {0}$. – copper.hat Oct 22 '20 at 20:41
  • If you find a continuous and bijective map $g$ from $\mathbb{R^2}$ to $\mathbb{R^2}$ it follows, for the universal property of quotient, that $\exists \bar g: \mathbb{R^2} /D^2 \to \mathbb{R^2}$ continuous and bijective whose inverse $\bar \pi$ is continuous. Have you ever seen the homeomorphism with $S^1$? In any case you cold consider the isomorphism between $\mathbb{R^2}$
    and $\mathbb{C}$     – Vajra Oct 22 '20 at 20:42

1 Answers1

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Define $$f : \mathbb R^2 \to \mathbb R^2, f(p) = \begin{cases} 0 & \lVert p \rVert \le 1 \\ \frac{\lVert p \rVert -1}{\lVert p \rVert} p & \lVert p \rVert \ge 1 \end{cases}$$ This is a continuos map such that $f^{-1}(0) = D^2$ and $f^{-1}(q) = \{ \frac{\lVert q \rVert +1}{\lVert q \rVert}q \}$ for $q \ne 0$. In particular $f$ is a surjection. Moreover, $f$ is closed map: Let $A \subset \mathbb R^2$ be closed. and $(q_n)$ be a sequence in $f(A)$ converging to some $q \in \mathbb R^2$. Choose $p_n \in A$ such that $f(p_n) = q_n$. If infinitely many $p_n \in D^2$, we find a subsequence $(p_{n_k})$ converging to some $p \in D^2$. Sinced $A$ is closed, $p \in A$. Then $q = \lim_n q_n = \lim_k q_{n_l} = \lim_k f(p_{n_l}) = f(p) \in f(A)$. Now let only finitely many $p_n \in D^2$. Throwing these away, we may assume w.l.o.g that all $p_n \notin D^2$. Then $\lVert p_n \rVert -1 = \lVert f(p_n) \rVert = \lVert q_n \rVert \to \lVert q \rVert$. Thus $(p_n)$ is bounded and we find a subsequence $(p_{n_k})$ converging to some $p \in \mathbb R^2$. Clearly $p \in A$ and as above we see that $q \in f(A)$.

As a closed map, $f$ is a quotient map. It induces a bijection $h : \mathbb R^2/D^2 \to \mathbb R^2$ such that $h \circ r = f$ ($r : \mathbb R^2 \to \mathbb R^2/D^2$ is the quotient map). This $h$ is automatically a homeomorphism. See When is a space homeomorphic to a quotient space?

Paul Frost
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