Show $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$ it bjective and continuous and its inverse is continuous I'm working through Renzo's Math 490 Introduction to Topology, http://www.math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf, and  on page 16, Example 1.8.4, it states:  

that a topologist cannot tell the difference between a circle $S^1
 =\{(x,y) \in \mathbb{R}^2\ |\ x^2+y^2=1\}$ and a square $T=\{(x,y)\in \mathbb{R}^2 \ |\ |x|+|y|=1\}$ as there is a function $f: S^1\rightarrow
 T$ defined by $f(x,y)=(\frac{x}{|x|+|y|},\frac{y}{|x|+|y|})$.

The book states that is is continuous and bijective, and that its inverse, $f^{-1}(x,y)=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$ is also continuous.  
How do I know the function is bijective and continuous, and that the inverse is continuous?
 A: Does this picture make it clear?

A: Both are given by 
$$f_i(x)=\frac{1}{\Vert x \Vert_i }x,$$
where $\Vert \cdot \Vert_i $ is the $l_1$ or $l_2$ norm. You can now use that any norm on a finite-dimensional vector space is equivalent, or simply the fact that those are simple norms with simple estimates.
In more detail on those estimates, let $\Vert x \Vert_1=\sum\limits_{j=1}^n \vert x_j \vert$ and $\Vert x \Vert_2=\sqrt{\sum\limits_{j=1}^n x_j^2}$. We have the following chain of inequalities, which are straightforward:
$$\Vert x \Vert _2 \leq \sqrt{n} \Vert x\Vert_{\infty} ,$$
$$\Vert x\Vert_{\infty} \leq \Vert x \Vert_1 \leq n\Vert x\Vert_2 .$$
Therefore, the map $x \mapsto \Vert x \Vert_i $ is continuous (since we "do not know" that all topologies are equivalent, we are assuming that we are given the topology induced by $\Vert x \Vert_2$, which is the standard norm on $\mathbb{R}^n$).
But note now that $f_i=m \circ(Id \times (\iota  \circ \Vert \cdot \Vert_i)),$where $\iota: \mathbb{R}^* \to \mathbb{R}$ is $\iota(r)=1/r$, and composition of continuous maps is continuous.
A: Injectivity: $$f(x,y) = f(a,b)\Rightarrow \dfrac{x^2+y^2}{(|x|+|y|)^2} = \dfrac{a^2+b^2}{(|a|+|b|)^2}.$$
But $x^2+y^2=a^2+b^2=1$, so $|xy| = |ab|$ or $x^2y^2 = a^2b^2\Rightarrow a^2+b^2-x^2-\dfrac{a^2b^2}{x^2}=0\Rightarrow (x^2-a^2)(x^2-b^2)=0,$ when $x\neq 0.$ If $|x|=|a|$, then $|y| = |b|$ and $\dfrac{x}{|x|+|y|} = \dfrac{x}{|a|+|b|}$. So it follows that $x=a$. If $|x| = |b|$, then similarly $x=a$, again. 
Surjectivity: Given $(x,y)\in T$, you can see that $f(\dfrac{x}{\sqrt{x^2+(1-x)^2}}, \dfrac{1-x}{\sqrt{x^2+(1-x)^2}}) = (x,y)$. 
Finally, the open mapping theorem implies that $f^{-1}$ is continuous. 
