Is this function one-one and onto? Consider $A=\{(x,y)\in \Bbb{R}^2 : x+y \neq -1\}$
Define $f:A \to \Bbb{R}^2$ by $$f(x,y)=\Big(\frac{y}{1+x+y},\frac{x}{1+x+y}\Big)$$
Then is it one-one on A and is it onto?
The jacobian of f has non zero determinant on A, therefore inverse function theorem assures that $f$ is locally one-one, but is it globally one-one on A? 
And it looks onto if we think each component to be similar to function $1/x$.
 A: Onto is pretty much answered in comments, so here is one-one.  If $f(x_1,y_1)=f(x_2,y_2)$ then
$$\frac{x_1}{1+x_1+y_1}=\frac{x_2}{1+x_2+y_2}\ ,\quad
  \frac{y_1}{1+x_1+y_1}=\frac{y_2}{1+x_2+y_2}\ .$$
Adding the equations,
$$\frac{x_1+y_1}{1+x_1+y_1}=\frac{x_2+y_2}{1+x_2+y_2}\ ;$$
multiplying out the denominators and noting that $(x_1+y_1)(x_2+y_2)$ can be cancelled,
$$x_1+y_1=x_2+y_2\ .$$
Substituting this back into the first pair of equations gives $x_1=x_2$ and $y_1=y_2$.
A: Suppose $f(x,y)=(a,b).$ Then as Thomas Andrews noted, $a+b-1 \neq 0.$ However we may with this condition solve for $x,y$ in terms of $a,b$ and get
$$x=\frac{b}{a+b-1},\ \ y=\frac{a}{a+b-1},$$ so it appears $f$ is injective.
Note injectivity here follows since there is only one pair $(x,y)$ for each pair $(a,b)$ in the range $\mathbb{R}^2 \setminus \{x+y-1=0\}.$
A: I think $f$ is one-one but $not$ onto.
Let $(a,b)\in \mathbb{R}^2$. Then for $f$ being onto we must have $(x,y)\in \mathbb{R}^2$ such that $f(x,y)=(a,b)$. If you will solve this condition then you will get that $x=\frac{b}{1-(a+b)},y=\frac{a}{1-(a+b)}$. Thus you can get all the values in $\mathbb{R}^2$ except those on the line $x+y=1$. You can then check that there are indeed $no$ such points in the plane whose function value lies on that line. Thus $f$ is not onto.
For seeing that $f$ is one-one, suppose $f(x,y)=f(p,q)$. Then by solving the corresponding condition you will get that $\frac xy=\frac pq$. Thus $f$ can fail to be one-one only on the line $y=mx, m\in \mathbb{R}$. But then if you will calculate the value of the function on $y=mx$, then you will see that $x=p$ and hence $y=q$, thus confirming that $f$ is one-one.
