Mixing
Image of smooth manifold is a submanifold
$begingroup$
It's know that if $M$ is a compact, smooth manifold of dimension $n$ and the map $f: M to mathbb{R^m}$ is injective, smooth, $n le m$ and $Jf(a)$, the Jacobian, has rank $n$ for every $a in M$, then $f(M)$ is a submanifold of $mathbb{R^m}$. I'm trying to think of a counterexample to this statement if we suppose $M$ is not compact, but haven't gotten anywhere. Can anyone offer a hint?
analysis differential-topology
asked Oct 4 '12 at 21:04
TylTyl
211
$endgroup$
add a comment |
$begingroup$
It's know that if $M$ is a compact, smooth manifold of dimension $n$ and the map $f: M to mathbb{R^m}$ is injective, smooth, $n le m$ and $Jf(a)$, the Jacobian, has rank $n$ for every $a in M$, then $f(M)$ is a submanifold of $mathbb{R^m}$. I'm trying to think of a counterexample to this statement if we suppose $M$ is not compact, but haven't gotten anywhere. Can anyone offer a hint?
analysis differential-topology
asked Oct 4 '12 at 21:04
TylTyl
211
$endgroup$
2
$begingroup$
Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18
add a comment |
$begingroup$
It's know that if $M$ is a compact, smooth manifold of dimension $n$ and the map $f: M to mathbb{R^m}$ is injective, smooth, $n le m$ and $Jf(a)$, the Jacobian, has rank $n$ for every $a in M$, then $f(M)$ is a submanifold of $mathbb{R^m}$. I'm trying to think of a counterexample to this statement if we suppose $M$ is not compact, but haven't gotten anywhere. Can anyone offer a hint?
analysis differential-topology
asked Oct 4 '12 at 21:04
TylTyl
211
$endgroup$
It's know that if $M$ is a compact, smooth manifold of dimension $n$ and the map $f: M to mathbb{R^m}$ is injective, smooth, $n le m$ and $Jf(a)$, the Jacobian, has rank $n$ for every $a in M$, then $f(M)$ is a submanifold of $mathbb{R^m}$. I'm trying to think of a counterexample to this statement if we suppose $M$ is not compact, but haven't gotten anywhere. Can anyone offer a hint?
analysis differential-topology
analysis differential-topology
asked Oct 4 '12 at 21:04
TylTyl
211
asked Oct 4 '12 at 21:04
TylTyl
211
asked Oct 4 '12 at 21:04
TylTyl
211
asked Oct 4 '12 at 21:04
TylTyl
211
asked Oct 4 '12 at 21:04
TylTyl
211
211
2
$begingroup$
Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18
add a comment |
2
$begingroup$
Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18
2
2
$begingroup$
Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18
add a comment |
1 Answer
1
active
oldest
votes
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Let $gcolon mathbb{R} to mathbb{T}^2= mathbb{R}^2/mathbb{Z}^2$ be given by $tmapsto [(t,alpha t)]$, where $alpha$ is any irrational number. The image of $g$ is a dense subset of $mathbb{T}^2$ but $g$ is not surjective. Hence, $mathrm{Im}(g)$ cannot be a submanifold of $mathbb{T}^2$. Compose this map with the embedding of the torus into $mathbb{R}^3$.
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
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add a comment |
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1 Answer
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$begingroup$
Let $gcolon mathbb{R} to mathbb{T}^2= mathbb{R}^2/mathbb{Z}^2$ be given by $tmapsto [(t,alpha t)]$, where $alpha$ is any irrational number. The image of $g$ is a dense subset of $mathbb{T}^2$ but $g$ is not surjective. Hence, $mathrm{Im}(g)$ cannot be a submanifold of $mathbb{T}^2$. Compose this map with the embedding of the torus into $mathbb{R}^3$.
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
$endgroup$
add a comment |
$begingroup$
Let $gcolon mathbb{R} to mathbb{T}^2= mathbb{R}^2/mathbb{Z}^2$ be given by $tmapsto [(t,alpha t)]$, where $alpha$ is any irrational number. The image of $g$ is a dense subset of $mathbb{T}^2$ but $g$ is not surjective. Hence, $mathrm{Im}(g)$ cannot be a submanifold of $mathbb{T}^2$. Compose this map with the embedding of the torus into $mathbb{R}^3$.
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
$endgroup$
add a comment |
$begingroup$
Let $gcolon mathbb{R} to mathbb{T}^2= mathbb{R}^2/mathbb{Z}^2$ be given by $tmapsto [(t,alpha t)]$, where $alpha$ is any irrational number. The image of $g$ is a dense subset of $mathbb{T}^2$ but $g$ is not surjective. Hence, $mathrm{Im}(g)$ cannot be a submanifold of $mathbb{T}^2$. Compose this map with the embedding of the torus into $mathbb{R}^3$.
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
$endgroup$
Let $gcolon mathbb{R} to mathbb{T}^2= mathbb{R}^2/mathbb{Z}^2$ be given by $tmapsto [(t,alpha t)]$, where $alpha$ is any irrational number. The image of $g$ is a dense subset of $mathbb{T}^2$ but $g$ is not surjective. Hence, $mathrm{Im}(g)$ cannot be a submanifold of $mathbb{T}^2$. Compose this map with the embedding of the torus into $mathbb{R}^3$.
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
answered Sep 20 '18 at 12:54
deepfloedeepfloe
714
714
add a comment |
add a comment |
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Take an open interval and map it into the plane such that it forms a figure "6". Focus on that non-manifold point on the 6 and think about how you can make something like that and still have an injective mapping.
$endgroup$
– yasmar
Oct 4 '12 at 21:15
$begingroup$
You can't do that with a proper map en.wikipedia.org/wiki/Proper_map
$endgroup$
– yasmar
Oct 4 '12 at 21:18