Example 4, Sec. 22, in Munkres' TOPOLOGY, 2nd ed: How is this quotient space homeomorphic with $S^2$?

Question

Here is Example 4, Sec. 22, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be the closed unit ball $$ \left\{ ( x, y) \lvert \ x^2 + y^2 \leq 1 \ \right\} $$ in $\mathbb{R}^2$, and let $X^*$ be the partition of $X$ consisting of all the one-point sets $\{ (x , y ) \}$ for which $x^2 + y^2 < 1$, along with the set $S^1 = \left\{ (x , y) \ \vert \ x^2 + y^2 = 1 \ \right\}$. Typical saturated open sets in $X$ are pictured by the shaded regions in Figure 22.4. One can show that $X^*$ is homeomorphic with the subspace of $\mathbb{R}^3$ called the unit-$2$-sphere, defined by $$ S^2 = \left\{ \ (x, y, z) \ \vert \ x^2+y^2+z^2 = 1 \ \right\}. $$

Now my question is, how is the space $X^*$ homeomorphic with the subspace $S^2$ of $\mathbb{R}^3$?

I'm sorry but I have not been able to make much sense of Figure 22.4 either; so can someone please also elaborate this to me?

Here are some relevant definitions.

Quotient Map:

Let $X$ and $Y$ be topological spaces; let $p \colon X \to Y$ be a surjective map. The map $p$ is said to be a quotient map provided a subset $U$ of $Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$. . . .

Saturated Sets:

We say that a subset $C$ of $X$ is saturated (with respect to the surjective map $p \colon X \to Y$) if $C$ contains every set $p^{-1} \left( \ \{ \ y \ \} \ \right)$ that it intersects. Thus $C$ is saturated if it equals the complete inverse image of a subset of $Y$. . . .

Quotient Topology:

If $X$ is a [topological] space and $A$ is a set and if $p \colon X \to A$ is a surjective map, then there exists exactly one topology $\mathscr{T}$ on $A$ relative to which $p$ is a quotient map; it is called the quotient topology induced by $p$.

The topology $\mathscr{T}$ is of course defined by letting it consist of those subsets $U$ of $A$ such that $p^{-1}(U)$ is open in $X$. . . .

Quotient Space:

Let $X$ be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p \colon X \to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology induced by $p$, the space $X^*$ is called a quotient space of $X$.

Finally, here is the image of Page 139 of Munkres. I would appreciate a detailed answer rather than a mere hint.

Answer 1

We say that $a\sim b$ for $a,b\in X$ iff $a,b$ belong to the same element from the considered partition. Therefore $$a\sim b\iff a=b\text{ or }a,b\in S^1. $$ This is an equivalence relation and $X^*$ is the set of the equivalence classes. Let $p\colon X\to X^*$ be a projection $p(x)=[x]_\sim$.

Let $f\colon X\to S^2$ be 'wrapping a sphere in a unit ball', that is $$f(r\cos\alpha,r\sin\alpha)=(\sin\beta\cos\alpha,\sin\beta\sin\alpha,\cos\beta),\text{ where }\beta = \pi r$$ (spherical coordinate system). Then $$f(a)=f(b)\iff a=b\text{ or }a,b\in S^1\iff a\sim b.\tag{1}\iff p(a)=p(b).$$ Therefore $f$ can be factorised by $X^*$, that is there exists a unique $f^*\colon X^*\to S^2$ such that $f^*\circ p = f$.

• From (1) we have $f^*$ is injection.
• Since $f$ is onto then $f^*$ is onto.
• Therefore $f^*$ is a bijection. We'll show that $f^*$ is a homeomorphism.

Fact $g\colon X^*\to Y$ is continuous iff $g\circ p$ is continuous.

Proof. From the very definition of the topology on $X^*$

Corollary $f^*$ is continuous.

It remains to show that $f^*$ is open. It's equivalent to show that $f(U)$ is open in $S^2$ for any open and saturated from $X$. Each such set is a sum of some sets from the family $$\mathcal B=\mathcal B_1\cup\{R_r:0<r<1\}\text{ where } \mathcal B_1=\{U\subset X\setminus S^1:U\text{ is open}\}\\\text{ and }R_r:=\{(x,y):r<x^2+y^2\leq 1\}.$$

• If $U\in\mathcal B$ then the openess of the image $f(U)$ can be shown analysing the derivative $Df$ of $f$.
• The set $f(R_r)=\{(x,y,z)\in S^2: z>\cos(\pi\cdot r)\}$ is open.

Answer 2

You can brute force this using basic properties of compact spaces vis á vis quotient maps. The map $\varphi:X\to S^2:$

$(x,y)\mapsto \left(\frac{1-\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}x,\frac{1-\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}y,2\sqrt{x^2+y^2}-1\right)$

is continuous (even at $(x,y)=(0,0)$, if we define $\varphi$ to be equal to zero there) and surjective. Since $X$ and $S^2$ are compact, $\varphi$ is a quotient map.

Now, $\varphi$ respects the identifications made by the quotient map $p$ that induces $X^*.$ That is, if $\|\vec x\|=1,$ then $\varphi$ sends $\vec x$ to $(0,0,1)$ whereas $\varphi$ is injective on the open unit disk $B(0,1).$ Thus, we get a continuous $\overline \varphi:X^*\to S^2$ that satisfies $\overline \varphi\circ p=\varphi.$ But $p$ also respects the identifications made by $\varphi$ so we get a continuous map going in the other direction $\overline p:X^*\to S^2.$

To finish, show that $\overline p$ and $\overline \varphi$ are inverses.