Exercise 2.3.6 (The unit circle is homeomorphic to a square)

Proof. Let $C$ denote the unit circle and $S$ denote the open square. Consider the following mapping:

We verify the following properties of $f$:

• $f(C)\subseteq S$.
  Let nonzero $(x,y)\in C$, i.e., $x^2+y^2<1$ (in particular, $\left|x\right|<1$ and $\left|y\right|<1$). We then have

  $$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)<\frac{\sqrt{x^2+y^2}}{\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)<\frac{1}{\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)\le \left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right),$$

  i.e., $f(x,y)\in S$.

• $f$ is invertible.
  Define

  $$g:S\to C,\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)\mapsto \{\begin{array}{cc}\frac{\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}{\sqrt{x^2+y^2}}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right),\hfill & \text{if }\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)\ne \left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)\hfill \\ \\ \left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)\hfill & \text{else.}\hfill \end{array}$$

  Then it is easy to verify that $f\circ g=g\circ f=\mathrm{id}<!--\; FUNCTION\; APPLICATION\; -->$, i.e., $g$ is the inverse of $f$.

• $f$ is continuous.
  For $(x,y)\ne (0,0)$, $f$ must be continuous at $(x,y)$ in every component since exponentiation, addition, division and the max/min functions are continuous as well and any composition of continuous functions is again continuous.
  At point $(0,0)$ we also have continuity:

  $$\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right)\le \frac{\sqrt{x^2+y^2}}{\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)\le \frac{\sqrt{2}\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}{\max <!--\; FUNCTION\; APPLICATION\; -->\{|x|,|y|\}}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=\sqrt{2}\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)\stackrel{x,y\to 0}{\text{→}}\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right).$$

  A function continuous in all components is continuous, as desired.

• $f^{-1}$ is continuous.
  Follows by the same argument.

Thus, we have established a homomorphism between the open circle and the open square. □