Sufficient and necessary conditions for a change of coordinates to be locally invertible
Given a 2D coordinate change $(x, y) mapsto (u, v)$, what are the sufficient and necessary conditions for the map to be invertible on every local neighborhood? For instance, the map,
$$ u = x^3 quad,quad v = y $$
has a Jacobian determinant which vanishes on the line $x = 0$, but yet has an inverse everywhere (if you only care about continuity of the inverse map).
For the 1D case, this is satisfied for maps $x mapsto f(x)$ which are strictly increasing (or decreasing), which I suppose can be stated as $f'(x) geq 0$ for all $x$, and $f'(x) = 0$ for distinct points only. Does this kind of logic generalize to the 2D case?
differential-geometry coordinate-systems jacobian
asked Dec 9 at 8:25
Kevin Chen
11
add a comment |
Given a 2D coordinate change $(x, y) mapsto (u, v)$, what are the sufficient and necessary conditions for the map to be invertible on every local neighborhood? For instance, the map,
$$ u = x^3 quad,quad v = y $$
has a Jacobian determinant which vanishes on the line $x = 0$, but yet has an inverse everywhere (if you only care about continuity of the inverse map).
For the 1D case, this is satisfied for maps $x mapsto f(x)$ which are strictly increasing (or decreasing), which I suppose can be stated as $f'(x) geq 0$ for all $x$, and $f'(x) = 0$ for distinct points only. Does this kind of logic generalize to the 2D case?
differential-geometry coordinate-systems jacobian
asked Dec 9 at 8:25
Kevin Chen
11
add a comment |
Given a 2D coordinate change $(x, y) mapsto (u, v)$, what are the sufficient and necessary conditions for the map to be invertible on every local neighborhood? For instance, the map,
$$ u = x^3 quad,quad v = y $$
has a Jacobian determinant which vanishes on the line $x = 0$, but yet has an inverse everywhere (if you only care about continuity of the inverse map).
For the 1D case, this is satisfied for maps $x mapsto f(x)$ which are strictly increasing (or decreasing), which I suppose can be stated as $f'(x) geq 0$ for all $x$, and $f'(x) = 0$ for distinct points only. Does this kind of logic generalize to the 2D case?
differential-geometry coordinate-systems jacobian
asked Dec 9 at 8:25
Kevin Chen
11
Given a 2D coordinate change $(x, y) mapsto (u, v)$, what are the sufficient and necessary conditions for the map to be invertible on every local neighborhood? For instance, the map,
$$ u = x^3 quad,quad v = y $$
has a Jacobian determinant which vanishes on the line $x = 0$, but yet has an inverse everywhere (if you only care about continuity of the inverse map).
For the 1D case, this is satisfied for maps $x mapsto f(x)$ which are strictly increasing (or decreasing), which I suppose can be stated as $f'(x) geq 0$ for all $x$, and $f'(x) = 0$ for distinct points only. Does this kind of logic generalize to the 2D case?
differential-geometry coordinate-systems jacobian
differential-geometry coordinate-systems jacobian
asked Dec 9 at 8:25
Kevin Chen
11
asked Dec 9 at 8:25
Kevin Chen
11
asked Dec 9 at 8:25
Kevin Chen
11
asked Dec 9 at 8:25
Kevin Chen
11
asked Dec 9 at 8:25
Kevin Chen
11
11
add a comment |
add a comment |
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