How to prove a topological isomorphism?

Question

A topological isomorphism between two topological spaces is a bijective, bicontinuous map.

Let $C$ be the set of points lying on the unit circle in $\mathbb{R^2}$ and let $D$ be the set of points lying on the ellipse $x^2/4+y^2=1/4$.

We equip $C,D$ with the standard metric on $\mathbb{R^2}$

Let $\psi$ be given by $\psi:(x,y) \to \left(\frac{x}{\sqrt{x^2+y^2}} ,\frac{y}{\sqrt{x^2+y^2}}\right)$ for all $(x,y) \in D$

Prove that $\psi$ is a topological isomorphism between $D$ and $C$.

Firstly we note that $\psi:D \to C$ since $\left(\frac{x}{\sqrt{x^2+y^2}} \right)^2+\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2=1$.

Now is where I get stuck we want to prove two things.

1) Firstly the map is bijective: I can show injectivity I think but I don't know how to show it is surjective (picturing the deformation of the ellipse into the circle it is clear that it will be but I can't prove it rigorously.)

2) How to show it is bicontinuous (Since we are working with the standard metric on continuity in the topological sense is equivalent to continuity in the metric space so maybe $\epsilon-\delta$ or use open sets to show this?)

The problem is I don't know what the inverse map is so I am struggling to prove either of 1) or 2).

Could anybody shed some light on this problem for me?

Thanks!

Answer 1

In this case, one may avoid calcualtions by using some theory. You don't have to prove that the inverse is continuous since every continuous map from a compact space to a Hausdorff space is closed. Also, we have that $ψ$ is a closed topological embedding to $C$, and since $C$ has no proper subspace topologically isomorphic (or homeomorphic) to itself, $ψ$ has to be onto.

Answer 2

Given a point $(a, b)$ on the unit circle, we are essentially moving it towards the origin until it collides with some point on the ellipse. Thus, we are solving the system of equations given by: $$ \begin{cases} y = \frac{b}{a}x \\ x^2 + 4y^2 = 1 \end{cases} $$ This yields an analogus inverse map $\psi^{-1}\colon C \to D$ given by: $$ (a, b) \mapsto \left( \frac{a}{\sqrt{a^2 + 4b^2}}, \frac{b}{\sqrt{a^2 + 4b^2}} \right) $$