Prove that $det df(x)$ cannot vanish on an open set
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Let $f: R^2to R^2$ be a $C^1$ map such that $f^-1(y)$ is finite for all $yin R^2$. Prove that $det df(x)$ cannot vanish on an open set.
My thoughts:
Suppose the determinant vanishes on an open set. Then it vanishes locally on every disc in the set. Any disc is convex, so $f$ is constant on that disc by the corollary to Rudin's Theorem 9.19 (Mean Value Theorem). So $f$ is locally constant. Let $D$ be one of the discs, $xin D$. Then $f^-1(f(x))$ contains the entire disc $D$, a contradiction.
Is this argument valid?
calculus real-analysis multivariable-calculus derivatives
asked Aug 9 at 15:17
user557902851
407117
 |Â
show 2 more comments
up vote
2
down vote
favorite
Let $f: R^2to R^2$ be a $C^1$ map such that $f^-1(y)$ is finite for all $yin R^2$. Prove that $det df(x)$ cannot vanish on an open set.
My thoughts:
Suppose the determinant vanishes on an open set. Then it vanishes locally on every disc in the set. Any disc is convex, so $f$ is constant on that disc by the corollary to Rudin's Theorem 9.19 (Mean Value Theorem). So $f$ is locally constant. Let $D$ be one of the discs, $xin D$. Then $f^-1(f(x))$ contains the entire disc $D$, a contradiction.
Is this argument valid?
calculus real-analysis multivariable-calculus derivatives
asked Aug 9 at 15:17
user557902851
407117
If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
Another hint: Rank.
– user539887
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f: R^2to R^2$ be a $C^1$ map such that $f^-1(y)$ is finite for all $yin R^2$. Prove that $det df(x)$ cannot vanish on an open set.
My thoughts:
Suppose the determinant vanishes on an open set. Then it vanishes locally on every disc in the set. Any disc is convex, so $f$ is constant on that disc by the corollary to Rudin's Theorem 9.19 (Mean Value Theorem). So $f$ is locally constant. Let $D$ be one of the discs, $xin D$. Then $f^-1(f(x))$ contains the entire disc $D$, a contradiction.
Is this argument valid?
calculus real-analysis multivariable-calculus derivatives
asked Aug 9 at 15:17
user557902851
407117
Let $f: R^2to R^2$ be a $C^1$ map such that $f^-1(y)$ is finite for all $yin R^2$. Prove that $det df(x)$ cannot vanish on an open set.
My thoughts:
Suppose the determinant vanishes on an open set. Then it vanishes locally on every disc in the set. Any disc is convex, so $f$ is constant on that disc by the corollary to Rudin's Theorem 9.19 (Mean Value Theorem). So $f$ is locally constant. Let $D$ be one of the discs, $xin D$. Then $f^-1(f(x))$ contains the entire disc $D$, a contradiction.
Is this argument valid?
calculus real-analysis multivariable-calculus derivatives
asked Aug 9 at 15:17
user557902851
407117
asked Aug 9 at 15:17
user557902851
407117
asked Aug 9 at 15:17
user557902851
407117
asked Aug 9 at 15:17
user557902851
407117
407117
If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
Another hint: Rank.
– user539887
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00
 |Â
show 2 more comments
If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
Another hint: Rank.
– user539887
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00
If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
Another hint: Rank.
– user539887
Aug 9 at 15:36
Another hint: Rank.
– user539887
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00
 |Â
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If I understand you well, you claim that if $detdf$ vanishes on an open set then $f$ is constant on that set. This is not true: take the projection on an axis. A hint: you have never used the fact that the inverse image of any point is finite
– user539887
Aug 9 at 15:35
Another hint: Rank.
– user539887
Aug 9 at 15:36
I can't validate your argument but I can propose an other idea: Take $xin mathbbR^2$ and $y=f(x)$. For $x+delta xin mathbbR^2$, there is $delta y$ such that $y+delta y = f(x+delta x)$. We have : $f(x+delta x) = f(x) + nabla_x fcdot delta x + o(delta x)$ Hence $delta y = nabla_x fcdot delta x + o(delta x)$. If $det nabla_x f$ vanishes, the linear system $delta y = nabla_x fcdot delta x$ has no solution or an infinite solutions which contradict the assumption on $f$. One problem of this solution is to take into account the remainder terms $o(delta x)$ ...
– Smilia
Aug 9 at 15:36
@user539887 I see, I tried to use Rudin's clam that if $df=0$ on a convex open set, then $f$ is constant. But $det df=0$ does not imply $df=0$. I think I did use the fact that the inverse image of any point is finite: I assumed that $df$ vanishes on an open set and found a point whose inverse image is infinite. But anyway this argument is invalid I guess.
– user557902851
Aug 9 at 15:48
If $detdf=0$ on some open connected set, then either the rank of $df$ is constantly equal to zero (then, by the Constant rank theorem the function is constant on some nbhd of any point), or there is a point where the rank is $1$. But then the rank is $1$ on some nbhd of that point, and by that same theorem the preimage is locally the graph of a $C^1$ (scalar) function.
– user539887
Aug 9 at 16:00