Sachchidanand Prasad

13-07-2023

Problem: Let $r_1, r_2 $ be two positive real numbers and let $D_{r_1}$ and $D_{r_2}$ be the closed discs in $\mathbb{R} ^2$ given by \begin{align*} D_{r_1} & = \left\{ (x,y) \in \mathbb{R} ^2 : x^2 + y^2 \leq r_1 \right\} \text { and } \\ D_{r_2} & = \left\{ (x,y) \in \mathbb{R} ^2 : x^2 + y^2 \leq r_2 \right\} . \end{align*} Prove that $D_{r_1}$ is homeomorphic to $D_{r_2}$, where $D_{r_1}$ and $D_{r_2}$ have their subspace topology from $\mathbb{R} ^2$.
Solution: Note that we need to shrink or expand the disc. We know that any two closed intervals are homeomorphic, in particular, \[ \left[ 0, r_1 \right] \xrightarrow{\cong} \left[ 0, r_2 \right] , ~ x \mapsto x \cdot \frac{r_2}{r_1}. \] In a similar way, we can define a map between the discs which will give a homeomorphism. \begin{align*} f : D_{r_1} \to D_{r_2},~ \mathbf{x} = \left( x_1, x_2 \right) \mapsto \mathbf{x} \cdot \frac{r_2}{r_1}. \end{align*}
homemorphism between discs

Let us first check that the above map is well-defined, in the sense that if $\lVert \mathbf{x} \rVert \leq r_1$, then $\left\lVert f\left( \mathbf{x} \right) \right\rVert \leq r_2$. \begin{align*} \left\lVert f\left( \mathbf{x} \right) \right\rVert = \left\lVert \mathbf{x} \cdot \frac{r_2}{r_1} \right\rVert = \lVert \mathbf{x} \rVert \left\vert \frac{r_2}{r_1} \right\vert \leq r_1 \frac{r_2}{r_1} = r_2. \end{align*} Clearly, $f$ is continuous and injective. As \[ f\left( \mathbf{x} \right) = f\left( \mathbf{y} \right) \implies \mathbf{x} \cdot \frac{r_2}{r_1} = \mathbf{y} \cdot \frac{r_2}{r_1} \implies \mathbf{x} = \mathbf{y} . \] The inverse of $f$ is defined as \[ f^{-1} (\mathbf{y} ) = \mathbf{y} \cdot \frac{r_1}{r_2},\kern 0.5cm \mathbf{y} \in D_{r_2}, \] which is also continuous. Therefore, $f$ is a homeomorphism between the two discs.