Multiple choice question on $f:A\to\mathbb{R}^2, f(x,y)=({x\over 1+x+y},{y\over 1+x+y})$

Asked Aug 21 '14 at 10:13

Active Dec 15 '15 at 11:39

Viewed 127 times

$A=\{(x,y)\in\mathbb{R}^2: x+y\neq -1\}$

$f:A\to\mathbb{R}^2, f(x,y)=({x\over 1+x+y},{y\over 1+x+y})$,Then

I have calculated $J={(x+y+2)(y-x)\over (x+y+1)^2}$, It will vanish only when $y=x$ or $x+y=-2$ but clearly they are in $A^c$, so $1$ is true.

Yes, $f$ is infinitely differentiable which I understand from the expression but which result I have to use to conclude that or how?

$f$ is injective as Non vanishing Jacobian

$f$ is not surjective as $({1\over 2},1)$ has no pre-image in $A$ Shall be glad to have reply

edited Dec 15 '15 at 11:39

Kushal Bhuyan

asked Aug 21 '14 at 10:13

Myshkin

What is $f(-1,-2)$? Regarding $1$ what's $J$ supposed to be? The jacobian matrix is a $2\times 2$ matrix. – Git Gud Aug 21 '14 at 10:21
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Perhaps OP is conflating matrix and determinant. – Gerry Myerson Aug 21 '14 at 12:41
Oh Yes, I thought they are asking whether determinant of the Jacobian matrix vanishes, anyway the matrix also does not vanish on $A$ – Myshkin Aug 21 '14 at 12:44
@la Bella Noiseuse... can u show how u find jacobian matrix of this function – Dec 06 '15 at 01:03
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Possible duplicate of Injection and Bijection of the function $f(x,y)=(\frac{x}{1+x+y},\frac{y}{1+x+y}).$ – detnvvp Dec 15 '15 at 11:43

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