What is the homeomorphism between a disk and an ellipse?

Question

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 $E$?

Answer 1

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 $$ h(C)=E \mbox{ and } h^{-1}(E)=C. $$

Answer 2

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.