<s>[INST] Injectivity and Range of a function Let, $A=\{(x,y)\in \Bbb R^2:x+y\neq -1\}$
Define, $f:A\to \Bbb R^2$ by $f(x,y)=\left(\displaystyle \frac{x}{1+x+y},\frac{y}{1+x+y}\right)$
$(1)$Is $f$ injective on $A$?
$(2)$What is the Range of $f$?
I started with $f(x,y)=f(u,v)$, then tried to show $(x,y)=(u,v)$. But I could not do it. Can someone help me out for this.
Secondly for computing the range, 
$$\begin{align}
f(x,y)=(a,b)\\
 \implies \displaystyle \frac{x}{1+x+y}=a,\displaystyle \frac{y}{1+x+y}=b\\
\implies (1-a)x-ay=a\\
-bx+(1-b)y=b\\
\implies x=\displaystyle \frac{a}{1-(a+b)},y=\displaystyle \frac{b}{1-(a+b)}
\end{align}$$
So, $f(A)=\Bbb R^2-\{(a,b)\in \Bbb R^2:a+b=1\}$
Is the range correct? If yec can it be obtained in some other way?
 [/INST] To continue our discussion in the comments, pick any $(x,y), (u,v) \in A$.
Suppose $f(x,y) = f(u,v)$. Further suppose $x, y, v \neq 0$. (You have to prove the cases (i) $x = 0$, (ii) $y = 0$ and (iii) $v = 0$ later on) 
Then $$\frac{x}{1+x+y} = \frac{u}{1+u+v}\quad \text{and} \quad \frac{y}{1+x+y} = \frac{v}{1+u+v}$$
For $\frac{y}{1+x+y} \neq 0$, we can divide $\frac{x}{1+x+y}$ by $\frac{y}{1+x+y}$, which is $\frac{v}{1+u+v}$. Then
$$\frac{x}{1+x+y} \frac{1+x+y}{y} = \frac{u}{1+u+v}\frac{1+x+y}{y}=\frac{u}{1+u+v}\frac{1+u+v}{v}$$
$$\frac{x}{y} = \frac{u}{v}$$
For $y, v \neq 0$, we can divide the numerator and divisor by $y$ and $v$.
$$\frac{\frac{x}{y}}{\frac{1}{y}+\frac{x}{y}+1} = \frac{\frac{u}{v}}{\frac{1}{v}+\frac{u}{v}+1}= \frac{\frac{x}{y}}{\frac{1}{v}+\frac{x}{y}+1} $$
And we can cancel $\frac{x}{y}$ on both sides, then
$$\frac{1}{\frac{1}{y}+\frac{x}{y}+1}= \frac{1}{\frac{1}{v}+\frac{x}{y}+1} $$
then
$$\frac{1}{y}+\frac{x}{y}+1 = \frac{1}{v}+\frac{x}{y}+1$$
then $v = y$. Then I think you can show that $x =u$ from now.
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