Why $\mathbb{D}^2 / \mathbb{S}^1 = \mathbb{S}^2$?

Asked Dec 16 '21 at 12:48

Active Dec 16 '21 at 13:25

Viewed 38 times

Let $\mathbb{D}^2$ be the closed planar unit disk, $\mathbb{D}^2 = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 1\}$, and let $\mathbb{S}^1$ be the boundary circle, $\mathbb{S}^1 = \{(x, y) \in \mathbb{R}^2 \mid x^2 +y^2 = 1\}$.

Why $\sim$ such that $\mathbb{D}^2 / \mathbb{S}^1 = \mathbb{S}^2$?

$\mathbb{S}^2$ denotes the sphere in $\mathbb{R}^3$.

My attempt:

Define an equivalence relation $\thicksim$ whose equivalence class of a point $(x, y)$ is $\{(x,y)\}$ when $(x, y) \not\in \mathbb{S}^1$ and $(x,y) \thicksim (1,0)$ whenever $(x,y) \in \mathbb{S}^1$. Is $\mathbb{D}^2/ \sim = \mathbb{S}^2$?

Please help me.

edited Dec 16 '21 at 13:25

asked Dec 16 '21 at 12:48

I don't understand how the equivalence has to be defined on $[0,1] \times [0,1]$ as you're dealing with $\mathbb S^1$? – mathcounterexamples.net Dec 16 '21 at 12:58
This might be a good time to review our guidelines for how to ask a good question, with emphasis on providing context. Your question is just a problem statement with no context, and as you may have experienced, is likely to attract close votes. – Lee Mosher Dec 16 '21 at 13:22
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Did you ever blow up a balloon and close it with a knot? – Pedro Dec 16 '21 at 13:29
@PedroTamaroff This look like a sphere, but what is the proof? Is there any? – Dec 16 '21 at 13:39
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A way to proceed is to find a map $D^2\to S^2$ that is injective away from the boundary and maps the boundary to the North Pole of $S^2$. Can you think of any such map? – Pedro Dec 16 '21 at 13:40
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If you remove the boundary class from the quotient you’re left with the plane, essentially. Just as with the sphere. – Henno Brandsma Dec 16 '21 at 14:02

Why $\mathbb{D}^2 / \mathbb{S}^1 = \mathbb{S}^2$?

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