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Example in the proof of Sard Theorem

Inverse of regular value is a submanifold, Milnor's proofCalculate $deg(f)$A question regarding the proof of Hopf's theorem.Milnor's proof of the fundamental theorem of algebra (Topology from the Differentiable Viewpoint)Question about Milnor's proof of Sard's TheoremTopology of Isolated SingularitiesUnderstanding Milnor's proof of the fact that the preimage of a regular value is a manifoldSard's Theorem Using Integration?Sard's theorem and the null space of the higher dimensional manifold.On the nature of a submanifold

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1

$begingroup$

In the book Topology from the differential viewpoint - Milnor, there
is a proof of Sard Theorem : When $f:X^{n+m}rightarrow Y^n$ is a
smooth map where $X^N$ has $N$ dimension, then $f(C^1)$ has a
measure $0$ where $C^1 ={ xin X |df_x=0}$.

Here $C^1$ is a set of isolated points ? Note that $C^1$ is an
intersection of $(n+m)n$ equations $frac{partial f_i}{partial
x_j}=0$. Here $C^1$ can be a curve or a surface ?

differential-topology

share|cite|improve this question

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

$endgroup$

add a comment | 

1

1

$begingroup$

In the book Topology from the differential viewpoint - Milnor, there
is a proof of Sard Theorem : When $f:X^{n+m}rightarrow Y^n$ is a
smooth map where $X^N$ has $N$ dimension, then $f(C^1)$ has a
measure $0$ where $C^1 ={ xin X |df_x=0}$.

Here $C^1$ is a set of isolated points ? Note that $C^1$ is an
intersection of $(n+m)n$ equations $frac{partial f_i}{partial
x_j}=0$. Here $C^1$ can be a curve or a surface ?

differential-topology

share|cite|improve this question

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

$endgroup$

add a comment | 

$begingroup$

In the book Topology from the differential viewpoint - Milnor, there
is a proof of Sard Theorem : When $f:X^{n+m}rightarrow Y^n$ is a
smooth map where $X^N$ has $N$ dimension, then $f(C^1)$ has a
measure $0$ where $C^1 ={ xin X |df_x=0}$.

Here $C^1$ is a set of isolated points ? Note that $C^1$ is an
intersection of $(n+m)n$ equations $frac{partial f_i}{partial
x_j}=0$. Here $C^1$ can be a curve or a surface ?

differential-topology

share|cite|improve this question

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

$endgroup$

share|cite|improve this question

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

share|cite|improve this question

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

asked Feb 22 at 6:46

HK LeeHK Lee

14.1k52360

1 Answer
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1

$begingroup$

Not necessarily is $C_1$ a set of isolated points. Just consider the example of a constant function $f: mathbb{R}^2 to mathbb{R}$. It is $df_{(x,y)} equiv 0$, so $C^1 = mathbb{R}^2$, a surface. If you want $C^1$ to be a curve, consider again any constant function $f: mathbb{R} to mathbb{R}$, whose $C^1$ is $mathbb{R}$, a curve.

Another example is the function $g:mathbb{R}^2 to mathbb{R}$, $g(x,y) = (x-y)^2$, which satisfies $df_{(x,y)} = 0$ if and only if $x =y$. This means that for the function $g$, $C^1 = { (x,y)in mathbb{R}^2 : x= y}$, which is a curve in $mathbb{R}^2$.

share|cite|improve this answer

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

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add a comment | 

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1

$begingroup$

Not necessarily is $C_1$ a set of isolated points. Just consider the example of a constant function $f: mathbb{R}^2 to mathbb{R}$. It is $df_{(x,y)} equiv 0$, so $C^1 = mathbb{R}^2$, a surface. If you want $C^1$ to be a curve, consider again any constant function $f: mathbb{R} to mathbb{R}$, whose $C^1$ is $mathbb{R}$, a curve.

Another example is the function $g:mathbb{R}^2 to mathbb{R}$, $g(x,y) = (x-y)^2$, which satisfies $df_{(x,y)} = 0$ if and only if $x =y$. This means that for the function $g$, $C^1 = { (x,y)in mathbb{R}^2 : x= y}$, which is a curve in $mathbb{R}^2$.

share|cite|improve this answer

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

$endgroup$

add a comment | 

1

$begingroup$

Not necessarily is $C_1$ a set of isolated points. Just consider the example of a constant function $f: mathbb{R}^2 to mathbb{R}$. It is $df_{(x,y)} equiv 0$, so $C^1 = mathbb{R}^2$, a surface. If you want $C^1$ to be a curve, consider again any constant function $f: mathbb{R} to mathbb{R}$, whose $C^1$ is $mathbb{R}$, a curve.

Another example is the function $g:mathbb{R}^2 to mathbb{R}$, $g(x,y) = (x-y)^2$, which satisfies $df_{(x,y)} = 0$ if and only if $x =y$. This means that for the function $g$, $C^1 = { (x,y)in mathbb{R}^2 : x= y}$, which is a curve in $mathbb{R}^2$.

share|cite|improve this answer

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

$endgroup$

add a comment | 

$begingroup$

Not necessarily is $C_1$ a set of isolated points. Just consider the example of a constant function $f: mathbb{R}^2 to mathbb{R}$. It is $df_{(x,y)} equiv 0$, so $C^1 = mathbb{R}^2$, a surface. If you want $C^1$ to be a curve, consider again any constant function $f: mathbb{R} to mathbb{R}$, whose $C^1$ is $mathbb{R}$, a curve.

Another example is the function $g:mathbb{R}^2 to mathbb{R}$, $g(x,y) = (x-y)^2$, which satisfies $df_{(x,y)} = 0$ if and only if $x =y$. This means that for the function $g$, $C^1 = { (x,y)in mathbb{R}^2 : x= y}$, which is a curve in $mathbb{R}^2$.

share|cite|improve this answer

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

$endgroup$

share|cite|improve this answer

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261

answered Mar 13 at 23:47

jorge hidalgojorge hidalgo

261