<s>[INST] Show that a function $f(x,y)$ is bijective Show that a function is bijective
$ f: \mathbb{R}_{+} \times  \mathbb{R}_{+} \rightarrow \mathbb{R}_{+} \times  \mathbb{R}, f(x,y) := (x+y; \frac{1}{x} - \frac{1}{y}) $ 
I know I have to show that this function is injection and surjection.
My attempts:


*

*Injection
$ f(x,y)=f(a,b) \Rightarrow  (x,y)=(a,b) $
so $(x+y; \frac{1}{x} - \frac{1}{y})= (a+b; \frac{1}{a} - \frac{1}{b} ) \Rightarrow (x,y)=(a,b)$
$x+y=a+b$ and $\frac{1}{x} - \frac{1}{y} =  \frac{1}{a} - \frac{1}{b} $
I write this as $(x-a)=(b-y)$ and $\frac{1}{x} - \frac{1}{a} = \frac{1}{y}-\frac{1}{b} $ which leads to:
$xa=-yb$
That doesn't make much sense consider that i had to prove that $x=a$ and $b=y$

*Surjection
Following the definition
For every $ (a,b) \in \mathbb{R}_{+} \times  \mathbb{R}$ there is $ (x,y) \in \mathbb{R}_{+} \times  \mathbb{R}_{+} $ such that $f(x,y) = (a,b)$
I tried to extract $y$ from $x+y=a$ and put it in $ \frac{1}{x}-\frac{1}{y}=b$
This gives me
$ y= \frac{ba-2 \pm \sqrt{4+a^2b^2}}{2b}$
so that doesn't look good...
 [/INST] Injectivity: You have shown that $x-a=b-y$ and $\frac{a-x}{ax}=\frac{b-y}{by}$. If $x-a$ and $b-y$ is not zero, then you have $ax=-by$ (as you got). But this is impossible since $a,b,x,y>0$. Hence, the only possibility is $x-a=b-y=0$. That is $(x,y)=(a,b)$.
Surjectivity: Your approach is fine, but you should separate cases $b>0$, $b=0$ or $b<0$. 
$b=0$ is quite easy. You can do it yourself.
$b>0$. Then only $y=\frac{ba-2 + \sqrt{4+a^2b^2}}{2b}$ is positive. And it's easy to show that $y<a$, so $x=a-y>0$.
$b<0$. Both roots are positive. You have to choose $y=\frac{ba-2 - \sqrt{4+a^2b^2}}{2b}$ so that $x=a-y>0$.
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