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

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?

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.

• Thanks. For (1): what do you mean by stating that "the term saturated is hardly standard in this context"? Following your explanation, there are three typical kinds of saturated open sets in $X$: the ones like $V$ (without boundary), the ones like $U$ with the outer boundary, and the ones like $U$ without boundaries. Is it right? For (2): How to measure the latitude from the north pole? Specifically, what is the angle? Furthermore, why is $\alpha = \pi$ (instead of $\alpha = \pi/2$) at the south pole? Commented Feb 3, 2014 at 4:36
• (1a) While the term "saturated" is used elsewhere in math with a totally unrelated meaning, I've never seen it before used with this meaning in connection with the quotient topology. (1b) I'm not sure how far you can generalize about types of saturated sets, beyond this specific example. Think about the two possibilities for $p(U)$: either include the center point, or don't. They are both open sets. Because $p$ maps the entire circumference of the disk to the center of $U$, it follows that you can include the circumference or omit it. Commented Feb 3, 2014 at 13:58
• For the meaning of the latitude angle, it's probably best to go look at the wikipedia article en.wikipedia.org/wiki/Latitude, especially the second diagram en.wikipedia.org/wiki/…. The latitude is the angle $\phi$. However, you'll note that in that diagram, $\phi$ is measured up from the equator. That's standard, of course. I want an angle like $\phi$, but measured down from the north pole. So in the northern hemisphere, $\alpha=\pi/2-\phi$ (using radians, $\pi/2 = 90^\circ$), and in the southern hemisphere, $\alpha=\phi+\pi/2$ Commented Feb 3, 2014 at 14:01

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 $$0\leq t\leq 1$$.