Construct a homeomorphis between two spaces For a unit sphere, choose a meridian from north pole to south pole. Identify it to one single point. Show that the quotient space is homeomorphic to the unit sphere.
I know that unit sphere is homeomorphic to the complex plane plus the infinity point. So the original problem is converted in to prove extended plane is homeomorphic to the extended plane with one line identified in to one single point. But I don't know how to construct the accurate homeomorphism.
 A: Consider the quotient space of a square $[0,1]^2$ using the following identifications:


*

*$(0,y) \sim (1,y)$ for all $0 \leq y \leq 1$.

*$(0,0) \sim (x,0)$ for all $0 \leq x \leq 1$.

*$(0,1) \sim (x,1)$ for all $0 \leq x \leq 1$.


Performing only the first identification, you would get a cylinder. The second and third identifications then collapse the top and bottom circles of the cylinder to (let's say) the north pole and the south pole respectively, resulting in a sphere. You can make the identification rigorous by using spherical coordinates and construct the homeomorphism in such a way that your specific meridian is identified in the model with the edge $\{ (0, y) \, | \, y \in [0,1] \}$. By collapsing it to a point, the total effect of all the identifications done is the same as collapsing the whole boundary of the square to a single point. This space is readily seen to be homeomorphic to a sphere.
A: Let $D$ be the unit disk in the $xy$-plane $\mathbb{R}^2$ and $S^2$ the unit sphere in $\mathbb{R}^3$ (with spherical coordinates, so $S^2$ is the graph of $\rho=1$).  
Define the continuous function $f:D\to S^2$ by $$f(x,y)=
\begin{cases} \left(\rho=1, \phi = \pi(1-y)/2, \theta = 2\pi \frac{x+\sqrt{1-y^2}}{2\sqrt{1-y^2}} \right) & \text{if }y\neq \pm 1 \\
(1,\pi(1-y)/2, 0) & \text{if }y = \pm 1  \end{cases}.$$
Essentially, this map just zips together the left/western arc $A_1$ of the bounding circle of $D$ with the right/eastern arc $A_2$ of the bound circle of $D$: i.e $(x_1,y_1)\sim (x_2,y_2)$ iff $y_1=y_2$ and $|x_1|=|x_2|=\sqrt{1-y_1^2}$.  Moreover, these identified arcs go the meridian of $S^2$.
Notice that $f$ is not injective, so $f^{-1}$ is a relation but not a function.  Though also note that $f$ is injective on the interior of $D$.
Now define a continuous function $g:D\to S^2$ as follows, using polar coordinates:
$$ g(r, \theta)= (\rho = 1, \phi = \pi r, \theta=\theta).$$
The map $g$ sends the bounding circle of $D$ to the south pole of $S^2$.  So $g$, again, is not injective, though it is injective on the interior of $D$ as well.
Now look at the relation composition $$h=g\circ f^{-1}:S^2\to S^2.$$ This is actually a function. (Off the meridian of the domain $h$ is an injective function and the image of the complement of the meridian in the domain is the complement of the south pole in the codomain.  In fact, $h$ is a function on the meridian as well: given a point $m$ on the meridian ($m$ neither the north or south pole), then $f^{-1}(m)$ is two points which $g$ sends to the south pole of the codomain of $g$.) 
Clearly $h$ is continuous at points not on the meridian. And any basic open disk neighborhood of the south pole has as its preimage under $g$ an open annular collar around the unit circle of $D$, which maps to a neighborhood of the main meridian via $f$.  So $h$ is continuous. 
But $h$ merely identifies the meridian in its domain to a single point  (the south pole) in its codomain.
Using the above formulae, one can derive an explicit formulation of $h$ in terms of spherical coordinates. 

