Let $A=\{(x,y)\in\mathbb{R}^{2}:x+y\neq-1 \}$
Define $$f:A\rightarrow\mathbb{R}^{2}$$ by $$f(x,y)=(\frac{x}{1+x+y},\frac{y}{1+x+y}).$$ I have to prove that the the function $f$ is one one but not onto. It is clear that $f$ is continuously differentiable but Inverse map theorem gives locally injectiveness of $f.$ How to show that $f$ is one one but not onto?
For onto i think $(u,v)$ such that $u+v-1=0$ does not have pre image. Please help me. Thanks in advance.
Write out the condition for $f(x,y)=(A,B):$
$$\eqalign{ x&=A(x+y+1)\\ y&=B(x+y+1) }$$
$$\eqalign{ (A-1)x+Ay+A&=0\\ Bx+(B-1)y+B&=0 }$$
This is a linear system with determinant $1-A-B$ so it does not have a unique solution if and only if $A+B=1.$
$f$ is not onto because it never reaches the point (A=1/2,B=1/2).
$f$ is one-to-one because whenever the determinant of the system is 0 it is unsolvable: put $B=1-A,$ then
$$\eqalign{ (A-1)x+Ay+A&=0\\ (1-A)x-Ay+(1-A)&=0 }$$
The sum of these two equations gives $1=0.$
