Quotient space of closed unit ball and the unit 2-sphere $S^2$

Question

This is an example from Munkres's Topology (Example 4 in Section 22 titled "The Quotient Topology", 2nd edition).

Example 4: Let $X$ be the closed unit ball $$\{ x \times y \mid x^2 + y^2 \le 1\}$$ in $\mathbb{R}^2$, and let $X^{\ast}$ be the partition of $X$ consisting of all the one-point sets $\{ x \times y \}$ for which $x^2 + y^2 < 1$, along with the set $S^1= \{ x \times y \mid x^2 + y^2 = 1 \}$. Typical saturated open sets in $X$ are pictured by the shaded regions in the figure below. One can show that $X^{\ast}$ is homeomorphic with the subspace of $\mathbb{R}^3$ called the unit 2-sphere, defined by $$S^2 = \{ x \times y \times z \mid x^2 + y^2 + z^2 =1 \}.$$

I am confused about two points in the example.

Problem 1: About the saturated open sets in $X$. In the example, the typical ones are pictured by the shaded regions ($U, V$) in the figure. However, I am not sure whether the boundaries (visually, by picture) of $U$ and $V$ are included. Particularly, there are two boundaries for $U$. Are they both contained in $U$? And why?

Problem 2: How to show that $X^{\ast}$ is homeomorphic with $S^2$? What is the mapping between them?

The following is my understanding:

Solution to problem 1: For $V$, the boundary is not included. For $U$, the outer boundary is included, while the inner one not. Is this solution right?

Answer 1

You're right, Munkres has given a very confusing picture. (Also, the term saturated is hardly standard in this context --- I've never seen in used in other topology books.)

For (1): you are correct about $V$. For $U$, your answer is a possible correct answer, but it would also be OK not to include the outer boundary; in that case, the corresponding open set on the sphere would look like a punctured open disk, i.e., missing a point inside.

For (2): the idea is that we can take the upper hemisphere and stretch it over the entire sphere, collapsing the equator to a single point (the south pole). To describe this formally, let the disk on the left have radius $\pi$. Use polar coordinates, so a point in the disk is $(r,\theta)$. On the sphere, use coordinates $(\alpha,\beta)$ where $\alpha$ is the angle of latitude, but measured from the north pole (so $\alpha=0$ at the north pole, $\alpha=\pi$ (i.e., $180^\circ$) at the south pole), and $\beta$ is the angle of longitude.

To be quite explicit about the definition of $\alpha$: In the diagram latitude, the latitude is the angle ϕ. However, you'll note that ϕ is measured up from the equator. That's standard, of course. I want an angle like ϕ, but measured down from the north pole. So in the northern hemisphere, α=π/2−ϕ (using radians, π/2=90∘), and in the southern hemisphere, α=ϕ+π/2.

Now let $(r,\theta)$ map to the point with $\alpha=r, \beta=\theta$. When $r=\pi$, we have the entire circumference mapping to the south pole.

Answer 2

About (2), one possible map is $$f(tx,ty)=\left(\cos(1-t)\pi, x\sin(1-t)\pi,y\sin(1-t)\pi\right)$$ where $(x,y)\in \partial \mathbb D$ where $\mathbb D$ is the open unit disc, and $0\leq t\leq 1$.