Are $\left\{ (x,y) \in \mathbb{R^2} | |y| > x^2 , |y| < 10 \right\} $ & $\left\{ (x,y) \in \mathbb{R^2}| |y| > x^4 , |y| < 10 \right\} $ homeomorphic?

Question

Are spaces $A=\left\{ (x,y) \in \mathbb{R^2} | |y| > x^2 , |y| < 10 \right\} $ and $B=\left\{(x,y) \in \mathbb{R^2} | |y| > x^4 , |y| < 10 \right\} $ homeomorphic? My answer is that they are because if we construct a function $ f \colon ClA \to ClB $ $ f(x, x^2) =(x,x^4) $ then f would be a continuous bijection and as $ClA$ is compact f would be a homeomorphism. Thus the restriction of f to A is also a homeomorphism.

Answer 1

Let's disect those sets. Both of them consist of two connected components: when $y<0$ and when $y>0$. It is enough that we show that those components are pairwise homeomorphic. And since each upper half component is a reflection of the lower half then it is enough to show that $A=\{(x,y)\ |\ y>x^2; y< 10\}$ and $B=\{(x,y)\ |\ y>x^4; y< 10\}$ are homeomorphic.

To do that you have to realize that both $f(x)=x^2$ and $f(x)=x^4$ are convex functions. A bit of work has to be done to make sure that these conditions together with $y<10$ imply that both $A$ and $B$ are open and convex subsets of $\mathbb{R}^2$. And it is well known that every open and convex subset of $\mathbb{R}^n$ is homeomorphic to a $n$-ball.

Answer 2

You can see this directly. Note that for $(x,y) \in A$ we have $x^2 < 10$, i.e. $\lvert x \rvert < \sqrt{10}$ and for $(x,y) \in B$ we have $\lvert x \rvert < \sqrt[4]{10}$. Define $\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = \dfrac{x}{\sqrt[4]{10}}$. This is a linear homeomorphism which maps the open interval $(-\sqrt{10},\sqrt{10})$ onto $(-\sqrt[4]{10},\sqrt[4]{10})$.

Define $$f : A \to B, f(x,y) = \left(\phi(x),\text{sgn}(y)\left(\phi(x)^4 + \dfrac{10 - \phi(x)^4}{10 - x^2}(\lvert y \rvert - x^2)\right)\right) .$$ Note that the $\text{sgn}$-function is continuous because $y = 0$ is impossible. Geometrically the line segments between $(x,x^2)$ and $(x,10)$ are mapped linearly onto the line segments between $(\phi(x),\phi(x)^4)$ and $(\phi(x),10)$ and the line segments between $(x,-x^2)$ and $(x,-10)$ are mapped linearly onto the line segments between $(\phi(x),-\phi(x)^4)$ and $(\phi(x),-10)$.

An inverse for $f$ is given by $$g : B \to A, g(x,y) = \left(\phi^{-1}(x), \text{sgn}(y)\left(\phi^{-1}(x)^2 + \dfrac{10 - \phi^{-1}(x)^2}{10 - x^4}(\lvert y \rvert - x^4)\right) \right) .$$ If you want you can formally check that $g \circ f = id_A$ and $f \circ g = id_B$.