What is the homeomorphism between a disk and an ellipse? A disk/circle is defined by 
$$C = \{(x,y) \in \mathbb{R^2} : x^2 + y^2 \leq r^2\}$$
An ellipse is defined by
$$E = \{(x,y) \in \mathbb{R^2}: x^2/a^2 + y^2/b^2 \leq 1 \}$$
How can we define a function that takes $C$ to $E$?
 A: Define the map
$$
h:\mathbb{R}^2\to\mathbb{R}^2, \quad h(x,y)=r^{-1}(ax,by).
$$
It is clear that $h$ is an isomorphism, and 
$$
h^{-1}(x,y)=r(a^{-1}x,b^{-1}y).
$$
Let $(u,v)=h(x,y)\in E$, i.e. $u=r^{-1}ax,v=r^{-1}by$. We have: 
$$
\frac{u^2}{a^2}+\frac{v^2}{b^2}\le 1 \iff r^{-2}\frac{(ax)^2}{a^2}+r^{-2}\frac{(by)^2}{b^2}\le 1 \iff x^2+y^2\le r^2, 
$$
i.e.
$$
(x,y)=h^{-1}(u,v) \in C.
$$
Hence
$$
h(C)=E \mbox{ and } h^{-1}(E)=C.
$$
A: Let $a,b $ be the major and minor axes respectively, and assume WOLG that the ellipse is in the form $$\frac {x^2}{b^2}+\frac{y^2}{a^2}=1 $$. Then use
$h: (x,y) \rightarrow (x,  \frac {a}{b}(y))$ to get the circle:
$$ \frac {x^2}{b^2}+ \frac {(a/b)^2y^2}{(a^2)}=1 =\frac {x^2}{b^2}+ \frac {y^2}{b^2} $$
(actually, you can also stretch the minor axis to get a circle).
This is a homeomorphism because scaling/stretching is a homeomorphism. Or you may use the fact that $h^{-1}( x, (a/b)y) = (x, (b/a)y)$ and I think it is clear that each of $h$ and $h^{-1}$ are continuous. So you have a continuous function $h$ with continuous inverse.
