In the quotient topology $D^2/{S^1} \cong S^2 $ Let $X = D^2 = \{(x, y) \in R^2\ : x^2 + y^2 \le 1\}$ be the closed unit disc (in the standard topology). Identify $S^1$ with  the boundary of $D^2$. 
Now I have to prove that $$D^2/S^1\cong S^2$$
I am unable to find out the exact homeomorphism. Please help me.
Thanks.
 A: Use $(r,\theta)\mapsto (\cos\theta\sin\pi r,\sin\theta\sin\pi r,\cos\pi r)$ from the unit disk with polar coordinates to the unit sphere in $\Bbb{R}^3$. Clearly this is continuous, and when $r$ is 1, all theta values map to $(0,0,-1)$. Hence this is a well defined map from $D^2/S^1\to S^2$. Since cosine has a continuous inverse on $[0,\pi]$, the map 
$$(x,y,z)\mapsto \left(\frac{x\arccos z}{\pi\sin(\arccos z)},\frac{y\arccos z}{\pi\sin(\arccos z)}\right)$$
 is a continuous map except at $z=-1$, and $z=1$ since there $\arccos z$ is $\pi$ and $0$ respectively, which are zeros of the sine function. At $z=1$, this is a removable discontinuity, since $$\lim_{x\to 0}\frac{x}{\sin x} =1,$$
so we have the limit at $(0,0,1)$ is $$\lim_{(x,y)\to(0,0)}(x/\pi,y/\pi)=(0,0).$$
Hence this is a continuous map from $S^2\setminus\{(0,0,-1)\}$ to the interior of the unit disk whose inverse is the first map. Now we hope that  we can extend this to $S^2$ by sending the last point to the point $S^1$ of $D^2/S^1$. To do this, we just need to show that the limit of the function approaches the boundary of the disk regardless of how one approaches the point $(0,0,-1)$. We therefore compose with $x^2+y^2$ and show the composite approaches $1$ as we approach $(0,0,-1)$. The composite is
$$ (x,y,z)\mapsto (x^2+y^2)\left(\frac{\arccos z}{\pi\sin\arccos z}\right)^2.$$
Then $x^2+y^2=1-z^2$, so we can reduce the limit to
$$\lim_{z\to -1^+} \frac{(1-z^2)(\arccos z)^2}{\pi^2(\sin\arccos z)^2}.$$
Replacing $z$ with $\cos\theta$ as $\theta\to \pi$ from the left, to get
$$\lim_{\theta\to\pi^-} \frac{(\sin\theta)^2 \theta^2}{\pi^2(\sin\theta)^2}=\lim_{\theta\to\pi^-}\frac{\theta^2}{\pi^2}=1.$$
Hence the function can be extended to give a continuous inverse of our  original function. Thus these are explicit homeomorphisms between $D^2/S^1$ and $S^2$.
A: Let $\mathbb{R}^2\mapsto \mathbb{H}$ (the quaternions) be defined by $f((x,\,y)) = \exp(x\,i + y\,j)$ where $i$ and $j$ are the quaternion units. Then:
$$D^2/S^1 = \left\{\exp(x\,i + y\,j)\mid\; x^2 + y^2 \leq \pi\right\} = f(D^2)\tag{1}$$
where we've defined $D^2 = \{z\in\mathbb{H}\mid \;|z|\leq\pi\}$. To understand this, use the Rodrigues formula version of the Taylor series definition for $\exp$:
$$\exp(x\,i + y\,j)  = \cos(\sqrt{x^2+y^2}) + \operatorname{sinc}(\sqrt{x^2+y^2})\,(x\,i + y\,j)\tag{2}$$
where, naturally:
$$\operatorname{sinc}(u)=\left\{\begin{array}{ll}\frac{\sin u}{u};&u\neq 0\\1&u=0\end{array}\right.$$
Our map $f$ clearly identifies points for for which $\sqrt{x^2 +y^2} = \pi$ by equating them all to the quaternion $-1$ and is bijective on the interior of $D^2$. So then, thinking of concentric geometric circles of the form $C(r) = \{f(r\,(\cos\theta,\,\sin\theta) ) \mid\; \theta\in\mathbb{R}\}$ in $D^2$ and their images in $\mathbb{H}$, where they are also geometric circles, we see that our set in (1) is the topological suspension of a circle. Indeed, it's a geometric sphere in $\mathbb{H}$. Of course, one has to go on to prove that $f$ and its well defined inverse are continuous, but that is a fairly easy exercise with the Rodrigues formula.
