Let
\begin{align*}
D^{2}&=\left\{(x,y)\in\mathbb{R}^{2}:x^{2}+y^{2}\leq1\right\}\text{ and }
\\
S^{2}&=\left\{(x,y,z)\in\mathbb{R}^{3}:x^{2}+y^{2}+z^{2}=1\right\}
\end{align*}
be endowed with the respective subspace topologies as subspaces of $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ and with the respective metric topologies. Define an equivalence relation $R$ on $D^{2}$ by $(x,y)R(x,y)$ for all $(x,y)\in D^{2}$ and $(a,b)R(c,d)$ if and only if $a^{2}+b^{2}=c^{2}+d^{2}=1$. Show that $D^{2}/R$ in the quotient topology is homeomorphic to $S^{2}$.
A problem linked to the above is found at Describe how to obtain a sphere in $\mathbb R^3$ as a quotient of a closed disk in $\mathbb R^2$ however, the answers provided do not answer my questions which are stated below.
First question: How can we find a surjective function $s:D^{2}\longrightarrow S^{2}$ that is also continuous so that if $\pi:D^{2}\longrightarrow D^{2}/R$ is the quotient map, then there exists a homeomorphism $H:D^{2}/R\longrightarrow S^{2}$ such that $H\circ\pi=s$. This question uses a theorem stated below.
Suppose that $(X,\mathcal{T}_{X})$ and $(Y,\mathcal{T}_{Y})$ are topological spaces and $p:X\rightarrow Y$ is an onto map between topological spaces that is continuous. Let $\sim$ be the equivalence relation on $X$ $x_{1}\sim x_{2}\Leftrightarrow f(x_{1})=f(x_{2})$, and $\widetilde{p}:X\rightarrow X/\sim$ the `natural' map sending each point of $X$ to its equivalence class. Then there exists a homeomorphism $h:X/\sim\rightarrow Y$ such that $p=h\circ\widetilde{p}$.
Progress thus far: I was hinted to define a function below
\begin{align*}
s:D^{2}\longrightarrow S^{2}~,~(x,y)\mapsto\begin{cases}
\left(\frac{\sqrt{1-(2r-1)^{2}}}{r}x,\frac{\sqrt{1-(2r-1)^{2}}}{r}y,(2r-1)\right)~,~\text{ if }0<r\leq 1
\\
(0,0,-1)~,~\text{ if }r=0
\end{cases}
\end{align*}
Related to first question: How does one arrive at such a definition of a function $s$? What motivates one to define such an $s$?
Second question: How does one show that $s$ is continuous? Ideally I would like to start with an open set $V\subseteq S^{2}$ and show that $f^{-1}(V)$ is open, but this does not seem to work well with the definition of $s$ above. Then I tried to use that $s$ is continuous on $D^{2}$ if and only if $s$ is continuous at every point $\alpha\in D^{2}$. This would mean that for any open set $V\subseteq S^{2}$ such that $s(x)\in V$, I need to find an open set $U\subseteq D^{2}$ such that $s(U)\subseteq V$. But using the usual metric topolgies, $V=B(\beta,\epsilon)$ for some $\epsilon>0$ with $\beta\in S^{2}$, to define a $U$ such that $s(U)\subseteq V$ does not seem very clear with the definition of $s$ given above.
Any help or explanation would be much appreciated.
Best Answer
I don't like messing with formulae, so I like to take the higher view point: In $D^2$ we identify the boundary circle $S^1$ to a point. The open disk, so $D^2\setminus S^1$, is homeomorphic to $\Bbb R^2$. (for this it is easier to give an explicit map, e.g. $x \to \frac{\|x\|}{\|x\|+1}x$ from $\Bbb R^2$ to the open disk). So the resulting quotient space is a copy of $\Bbb R^2$ with an extra point added which makes it compact (as the quotient of a compact space is compact). So it's the one-point compactification of $\Bbb R^2$, which is essentially unique (this is a standard theorem in many text books).
And $S^2$ also is homeomorphic to the one-point compactification of $\Bbb R^2$ via the standard stereographic projection map.
And two spaces homeomorphic to the same space are homeomorphic to each other (to paraphrase Euclid).
We can make this explicit by composing the inverse of the above map between the open disk and the plane and the stereographic formula I found here, e.g. and then we get a rather ugly formula, for $(x,y) \in D^2\setminus S^1$ (so that $\sqrt{x^2+y^2}=\|(x,y)\|<1$):
$$f(x,y) = (\frac{2h_1(x,y)}{1+\|h(x,y)\|^2}, \frac{2h_2(x,y)}{1+\|h(x,y)\|^2}, \frac{-1 + \|h(x,y)\|^2}{1+\|h(x,y)\|^2})$$
where $$h(x,y)=(h_1(x,y), h_2(x,y)) = (\frac{\|(x,y)\|}{1-\|(x,y)\|}x, \frac{\|(x,y)\|}{1-\|(x,y)\|}y)$$
and sending all points $(x,y)$ that have norm $1$ to $(0,0,1)$. Then $f: D^2 \to S^2$ obeys $f(x,y) = f(u,v) \iff (x,y)\sim (u,v)$ etc. The continuity follows from the considerations in the first part.