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Let $X$ be a topological space and let $A ⊂ X$. Let $\sim$ be an equivalence relation on $X$ such that the equivalence classes are:

  • $A$ itself and
  • Singletons $\{x\}$ such that $x ∉ A$.

Then define $X/A$ to be the quotient space $X/{\sim}$. (i.e. collapse $A$ to a point)

Let $D^2$ denote the unit disk in $R^2$ and let $S^1$ denote its boundary.

What's a fairly well-known space, then, that $D^2/S^1$ is homeomorphic to? Is it $S^2$? And why?

user642796
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    Jimmy, do not do this. You just edited two of your questions, the other being this question, to ask a complete different question identical at both posts. You were totally invalidating the answers given on both counts. – Mark Fantini Jan 05 '15 at 02:44

3 Answers3

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Yes. $S^1$ (the boundary of $D^2$) is being contracted to a single point. And the quotient map restricted to the interior of $D^2$ is a homeomorphism.

You may also want to note that the open disc is homeomorphic to the plane $\mathbb R ^2$, which is homeomorphic to the deleted sphere $S^2\setminus \{p\}$ via the stereographic projection. We can identify the singleton $\{p\}$ with the contracted boundary of $D^2$.

Think of a gym bag with a drawstring lying flat on the floor. Then you lift it, pulling the drawstring closed.

Forever Mozart
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    So it is homeomorphic to $S^2$ then? –  Dec 12 '14 at 23:32
  • @Jimmy Yes it is. – Forever Mozart Dec 12 '14 at 23:34
  • Ok thank you. In a few sentences, how could I show that? "$S^1$ (the boundary of $D^2$) is being contracted to a single point. The quotient map restricted to the interior of $D^2$ is a homeomorphism." Could you help me go into a little more detail? –  Dec 12 '14 at 23:37
  • (+1) Fantastic intuitive answer! And here I was, misreading the question! (I was searching for a familiar space homeomorphic to $\Bbb R^2/S^1$, instead.) – Cameron Buie Dec 12 '14 at 23:40
  • @Jimmy The quotient $D^2/S^1$ collapses $S^1$ to a single point by definition. I updated my answer to show how the homeomorphism can be constructed. – Forever Mozart Dec 12 '14 at 23:48
  • Yes thank you very much for this explanation –  Dec 12 '14 at 23:51
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Yes, $D^2 / S^1$ is homeomorphic to $S^2$ and in general, $D^n / \partial D^n$ is homeomorphic to $S^n$. To see this intuitively, take a disc and place a sphere tangent to its center. Pull the boundary of the disc up to the equator so that the disc covers a hemisphere, then move the boundary along the surface of the sphere to the uncovered pole; at the pole, the boundary will have to be glued together.

To actually show homeomorphism, we of course need to construct a map. As a starting point, I would send the center of the disc to one pole and the boundary to another.

If you know about stereographic projection, you might be able to modify that somewhat to apply here.

Benjamin
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To add to the points already been made, one can think of $D^2$ as follows : $$D^2=\{ (tx_1, tx_2) \mid 0\leq t \leq 1, (x_1,x_2)\in S^{1}\}. $$ Now define a map $f:D^2\rightarrow S^2$ by $(tx_1, tx_2)\mapsto (ux_1, ux_2, 2t-1)$, where $u=(1-(2t-1)^2)^{1/2}$. Basically it sends the ray from $0$ to $(x_1, x_2)$ to the longitude that runs from the south pole $(0,0,−1)$ through the equatorial point $(x_1,x_2,0)$ to the north pole $(0, 0, 1)$.

Now the universal property of the quotient gives you a unique induced map $\widehat{f}:D^2/S^1\rightarrow S^2$ from compact to Hausdorff spaces.

mph
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