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?

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. 
