Finding an explicit homeomorphism from D^2 / S^1 to S^2 so I've seen a similar question from about 10 years ago on here, but I didn't see anything really alluding to the direction I was trying to write an explicit homeomorphism in, and I've just been struggling a little bit (I think it is the seeming asymmetry of the problem that might be confusing me). Here's what I've done so far;
$D^2 / S^1$ is defined as the unit disk, quotient out by the equivalence relation $~$ which is generated by identifying all points on the boundary of $D^2$ with each other. To map this space to the unit sphere $S^2$, I am thinking I should map;

*

*the center $(0,0)$ onto the north pole


*all circles with radius less    than some value $a$ onto circles on
the upper hemisphere of S^2


*the circle of radius $a$ onto the equator


*all circles with radius greater than $a$ (but less than 1) onto
circles on the lower hemisphere


*and finally mapping the equivalence class corresponding to the
boundary onto the south pole.
I have come up with the north hemisphere map (I think);
$ (x,y) \rightarrow (\frac{x}{a}, \frac{y}{a}, \sqrt{1 - \frac{x^2 + y^2}{a^2} } )$
This satisfies the equation of the sphere, and should be bijective (when the image is considered to be the upper hemisphere). I cannot for the life of me figure out the mapping onto the lower hemisphere though! can anyone please tell me what I am missing, as I am sure there should be some sort of reflection + translation business one can perform to obtain the proper map for the lower half of the sphere. Thanks heaps for reading.
 A: Your approach is absolutely correct. The idea is to define $f : D^2 \to S^2$ by mapping

*

*$0 \in D^2$ to the north pole $n$ of $S^2$

*each circle $S^1_t = \{ z \in D^2 \mid \lVert z \rVert = t \}$ with $0 < t < 1$ homeomorphically onto to the circle $S^2_t =  S^2 \cap (\mathbb R^2 \times \{1-2t\})$

*the circle $S^1 = S^1_1$ onto the south pole $s$ of $S^2$
The circle $S^2_t$ has radius $r_t = \sqrt{1 -(1-2t)^2} = 2\sqrt{t - t^2}$. From this "motivational construction" we get the explicit formula
$$f(z) = \begin{cases} (0,1) &  z = 0 \\
 \left(\frac{2\sqrt{\lVert z \rVert - \lVert z \rVert^2}}{\lVert z \rVert}z, 1 - 2\lVert z \rVert \right) & z \ne 0 \end{cases}$$
Continuity in all $z \ne 0$ is obvious. Continuity in $z = 0$ follows from
$$\lVert f(z) - f(0) \rVert^2 = 4(\lVert z \rVert - \lVert z \rVert^2) +4 \lVert z \rVert^2 = 4 \lVert z \rVert$$
which shows that $f(z) \to f(0)$ as $z \to 0$. By construction $f$ maps $D^2 \setminus S^1$ bijectively onto $S^2 \setminus \{s\}$. You can verify this formally: For $0 < \lVert z  \rVert, \lVert z'  \rVert < 1$ the equation $f(z) = f(z')$ implies $\lVert z  \rVert = \lVert z'  \rVert$ (compare second coordinates) and hence $z = z'$ (compare first coordinates where $\lVert z  \rVert = \lVert z'  \rVert$). Moreover, for $0 < t < 1$ each point $(w,1-2t) \in S^2_t$ (which means that $\lVert w \rVert^2 + (1-2t)^2 = 1$, i.e. $\lVert w \rVert = 2\sqrt{t - t^2}$) has the form $f(z)$ with $z = \frac{t}{2\sqrt{t - t^2}}w$; just note that $\lVert z \rVert = t$.
Since $f$ maps $S^1$ onto $s \in S^2$, we see that $f$ induces a homeomorphism
$$h : D^2/S^1 \to S^2, h([z]) = f(z).$$
