What space is this Homeomorphic to?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.