Munkres topology excercise show that the quotient space of a disk is homeomorphic to the unit sphere Find a homeomorphism between a quotient space of disk and sphere. I was looking at this example from munkres quotient topology section:Let $X$ be the closed unit ball $\{(x,y)|x^2+y^2\leq1\}$ in $\mathbb{R}^2$ and let $X*$ be the partition of $X$ consisting of one point sets $\{(x,y)\}$ for which $x^2+y^2<1$ along with the set $S^1=\{(x,y)|x^2+y^2=1\}$. One can show that $X*$ is homeomorphic with the unit two sphere $S^2=\{(x,y,z)|x^2+y^2+z^2=1\}$, a subspace of $\mathbb{R}^3$.
Here was my attempt at this problem Let $X*=\{\bigcup\{(x,y)|x^2+y^2=1\} \} \cup \{\{(x,y)\}|x^2+y^2<1\}$
My question is would this be a valid homeomorphism between $X*$ and $S^2$?
Define the function $f:X* \rightarrow S^2$ given by $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ where $0<x \leq 1$ and $0<y \leq 1$ and $f(\bigcup\{(x,y)|x^2+y^2=1\})=(0,0,1)$.
$f$ is injective since $f(\{(x,y)\}=f(\{w,z\}) \implies (\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)=\cos2\pi w \sin \pi z, \sin 2\pi w \sin  \pi z, \cos \pi z)\implies x=w,y=z \implies \{x,y)\}=\{(w,z)\}$.
$f$ is surjective since for any $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)\in S^2$ if $x=1,y=0$ then $f(\bigcup\{(x,y)|x^2+y^2=1\})=(0,0,1)$, otherwise $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$.
$X*$ is compact since it is the image of $S^1$ under the quotient map. $S^2$ is Hausdorff as a subspace of $\mathbb{R}^3$. Further more $f$ is continuous since if $g$ denotes the quotient map, $f \circ g:S^1 \rightarrow S^2$ is continuous since each component function in $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ of $(0,0,1)$ is continuous. Thus $f$ is a homeomorphism being a continuous bijection from a compact space to a Hausdorff space.
I am fairly confident that this is correct but is my restriction of the function $f$ mapping the equivalence class $\{(x,y)\}$ to $(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$ exactly correct, by restricting $x,y$ values to $0<x \leq 1$ and $0<y \leq 1$?
 A: Your definition does not work. You map $S^1$ to the north pole $N = (0,0,1)$ and for all other points $\{(x,y)\}$ of $X^*$ you define $f(\{(x,y)\})=(\cos2\pi x \sin \pi y, \sin 2\pi x \sin  \pi y, \cos \pi y)$.
These "other points" are characterized by
$$x^2 + y^2 < 1 \tag{1}$$
and not by $0 < x,y \le 1$. In fact $(1)$ allows that both $x,y$ can take the value $0$, but none of them can take the value $1$.
Anyway, $f$ is not injective. You have $f(\{(x,0)\}) = (0,0,-1)$ for $-1 < x < 1$.
Finally, whatever the definition of $f : X^* \to S^2$ looks like, you must verify the continuity with respect to the quotient topology on $X^*$. Your definition does not produce a continuous map. If it were, then $F = fq : D^2 \to S^2$ would be continuous, where $q : D^2 \to X^*$ is the quotient map But $\lim_{n \to \infty} F(1-\frac{1}{n},0) = (0,0,-1) \ne (0,0,1) = F(1,0)$.
So what can be done? Let us construct a map $F: D^2 \to S^2$ as follows:

*

*Map $(0,0)$ to the south pole $S = (0,0,-1)$ and $S^1$ to $N$.


*Map each circle $S^1_t = \{(x,y) \in D^2 \mid  \lVert (x,y) \rVert ^2 = x^2 + y^2 = t \}$ with $0 < t  < 1$ to the intersection circle $C(t)$ of $S^2$ with the plane $P(t) = \{(x,y,z) \mid z = 2t -1\}$ which has radius $r(t) = \sqrt{1 - (2t-1)^2}$.
That is, define
$$F(\xi)  = \begin{cases}\left(\frac{\sqrt{1 - (2\lVert \xi \rVert-1)^2}}{\lVert \xi \rVert} \xi,2\lVert \xi \rVert - 1\right) & \lVert \xi \rVert > 0 \\ S & \xi = (0,0) \end{cases}$$
Note that $F(\xi) = N$ for $\lVert \xi \rVert = 1$. The map $F$ is obviously continuous on $D^2 \setminus \{(0,0)\}$. For $\xi \ne (0,0)$ we have
$$\lVert F(\xi) - F(0,0) \rVert = \left\lVert \left(\frac{\sqrt{1 - (2\lVert \xi \rVert-1)^2}}{\lVert \xi \rVert} \xi,2\lVert \xi \rVert \right) \
\right\rVert = 4 \lVert \xi \rVert$$
which proves continuity in $(0,0)$.
Moreover $F$ is surjective. Clearly $S, N$ are in the image of $F$. Now let $(\eta,\tau) \in S^2 \setminus \{S,N\}$, where $\eta = (a,b)$ with $a,b \in \mathbb R$ and $-1 < \tau < 1$. Note that $\lVert \eta \rVert > 0$ and $\lVert \eta \rVert^2 + \tau^2 = 1$. Define $\xi =  \frac{1 +\tau}{2\sqrt{1 - \tau^2}}\eta$. Then $\lVert \xi \rVert =\frac{1+\tau}{2}$ ( so that $0  < \lVert \xi \rVert  < 1$) and we have $F(\xi) = (\eta,\tau)$.
The fibers $F^{-1}(\eta,\tau)$ are singletons for $(\eta,\tau) \ne N$ and $F^{-1}(N) = S^1$.
Thus $F$ gives us precisely your partition $X^*$.
A: More conceptually: letting $D= \{(x,y)\mid x^2 + y^2 \le 1\}$, $B = \{(x,y)\mid x^2 + y^2 < 1\}$, $S^1 = \{(x,y)\mid x^2 + y^2 = 1\}$, we see that the quotient $X^\ast$ of $D$ is just $D{/}S^1$ which is compact Hausdorff (as the quotient of the compact and $T_4$ space $D$, by identifying a single closed set to a point) and $X^\ast\setminus \{S^1\} \simeq B \simeq \Bbb R^2$ so that $X^\ast$ (by a standard theorem) is homeomorphic to the one-point (Aleksandrov) compactification of $\Bbb R^2$, which is just $S^2$ (stereographic projection etc).
No need really for an explicit function and we get the general fact that $D^n{/}S^{n-1} \simeq S^n$ for all $n$ in the same way.
