Show that f is a homeomorphism?

Question

I'm trying to show that $\vec{f}: D \to R, \vec{f}(r, \varphi)= \begin{pmatrix}r\cos(\varphi)\\ r\sin(\varphi)\end{pmatrix} $ is a homeomorphism, with $D = \{(r, \varphi) \in \mathbb{R^2}|r\in[1,2], \varphi \in [o, \frac{\pi}{2}]\}$ and on the other hand with $R = \{(x,y) \in \mathbb{R^2}|y\geq 0,x\geq 0, 1 \leq x^2+y^2 \leq 4\}$. Now, I'm not quite sure how I should go about this. Do I just find any function that assigns values to $x$ and $y$ in terms of $r$ and $\varphi$ and then show that continuity, bijection and continuity of the inverse are given? Any help is greatly appreciated since I'm fairly new to this. Best regards.

Answer 1

This really depends on which results about homeomorphisms you assume to be true, but keeping it quite simple, in the roum of what you know, I would attack the problem this way:

• Prove $\vec{f}$ is continuous (simple as $f$ is a product of elementary functions)
• Find an inverse for $\hat{f}$ (it won't even look ugly!)
• Prove that inverse is also continuous (also straight-forward)

Answer 2

Clearly $f$ is continuous.

Take $(x,y) \in R$ and assume that $f(r,\varphi)= (x,y)$ for some $(r, \varphi) \in D$. We get $$(r\cos\varphi, r\sin\varphi) = f(r, \varphi) = (x,y)$$ so it follows $r = \sqrt{x^2+y^2}$ and $\cos\varphi = \frac{x}{r}$ so $$\varphi= \arccos\frac{x}{\sqrt{x^2+y^2}}$$

After checking that indeed $(r, \varphi) \in D$ and $f(r,\varphi) = (x,y)$, we conclude that $f \circ g = \operatorname{id}_R$ where $g : R \to D$ is given by $$g(x,y) = \left(\sqrt{x^2+y^2}, \arccos\frac{x}{\sqrt{x^2+y^2}}\right)$$ which is a continuous function $R \to D$. Now also check that $g \circ f = \operatorname{id}_D$ to conclude that $f$ is bijective with inverse $f^{-1} =g$.

Hence $f$ is a homeomorphism.