Extending smooth maps from the punctured disk
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Suppose that I have a smooth map of manifolds $f: M^n rightarrow D^2backslash 0$ and $f$ is a submersion, i.e. a fiber bundle over $D^2 backslash 0$. When is it possible to find another smooth map of manifolds $g: N^n rightarrow D^2$ such that $M^n subset N^n$ and $g$ extends $f$? Ideally, I would also want that $N^n backslash M^n$ has codimension 1 in $N^n$.
Note that I am allowed to pick $N^n$, unlike in the following post Extending a smooth map
where $N^n$ is prescribed a priori.
If I don't care that $N$ and $g$ are smooth, then this is always possible. Just take the one point compactification $hat{M}$ of $M$ and extend the map $f$ continuously to compactification.
smooth-manifolds homotopy-theory
asked Dec 18 '18 at 16:56
user39598user39598
174213
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add a comment |
$begingroup$
Suppose that I have a smooth map of manifolds $f: M^n rightarrow D^2backslash 0$ and $f$ is a submersion, i.e. a fiber bundle over $D^2 backslash 0$. When is it possible to find another smooth map of manifolds $g: N^n rightarrow D^2$ such that $M^n subset N^n$ and $g$ extends $f$? Ideally, I would also want that $N^n backslash M^n$ has codimension 1 in $N^n$.
Note that I am allowed to pick $N^n$, unlike in the following post Extending a smooth map
where $N^n$ is prescribed a priori.
If I don't care that $N$ and $g$ are smooth, then this is always possible. Just take the one point compactification $hat{M}$ of $M$ and extend the map $f$ continuously to compactification.
smooth-manifolds homotopy-theory
asked Dec 18 '18 at 16:56
user39598user39598
174213
$endgroup$
$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
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– Lukas Geyer
Dec 18 '18 at 17:09
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Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
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This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17
add a comment |
$begingroup$
Suppose that I have a smooth map of manifolds $f: M^n rightarrow D^2backslash 0$ and $f$ is a submersion, i.e. a fiber bundle over $D^2 backslash 0$. When is it possible to find another smooth map of manifolds $g: N^n rightarrow D^2$ such that $M^n subset N^n$ and $g$ extends $f$? Ideally, I would also want that $N^n backslash M^n$ has codimension 1 in $N^n$.
Note that I am allowed to pick $N^n$, unlike in the following post Extending a smooth map
where $N^n$ is prescribed a priori.
If I don't care that $N$ and $g$ are smooth, then this is always possible. Just take the one point compactification $hat{M}$ of $M$ and extend the map $f$ continuously to compactification.
smooth-manifolds homotopy-theory
asked Dec 18 '18 at 16:56
user39598user39598
174213
$endgroup$
Suppose that I have a smooth map of manifolds $f: M^n rightarrow D^2backslash 0$ and $f$ is a submersion, i.e. a fiber bundle over $D^2 backslash 0$. When is it possible to find another smooth map of manifolds $g: N^n rightarrow D^2$ such that $M^n subset N^n$ and $g$ extends $f$? Ideally, I would also want that $N^n backslash M^n$ has codimension 1 in $N^n$.
Note that I am allowed to pick $N^n$, unlike in the following post Extending a smooth map
where $N^n$ is prescribed a priori.
If I don't care that $N$ and $g$ are smooth, then this is always possible. Just take the one point compactification $hat{M}$ of $M$ and extend the map $f$ continuously to compactification.
smooth-manifolds homotopy-theory
smooth-manifolds homotopy-theory
asked Dec 18 '18 at 16:56
user39598user39598
174213
asked Dec 18 '18 at 16:56
user39598user39598
174213
edited Dec 18 '18 at 17:11
user39598
asked Dec 18 '18 at 16:56
user39598user39598
174213
asked Dec 18 '18 at 16:56
user39598user39598
174213
asked Dec 18 '18 at 16:56
user39598user39598
174213
174213
$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:09
$begingroup$
Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
$begingroup$
This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17
add a comment |
$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:09
$begingroup$
Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
$begingroup$
This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17
$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:09
$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:09
$begingroup$
Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
$begingroup$
Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
$begingroup$
This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17
$begingroup$
This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17
add a comment |
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$begingroup$
First, not every submersion is a fiber bundle, so you probably want your maps to be surjective and proper. Also, the one-point compactification does not provide a continuous extension, as you can already see from the example where $M^n = D^2 setminus 0$ and $f$ is the identity. By the way, is it supposed to be $D^n$ or $D^2$ for the extension?
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:09
$begingroup$
Thanks! But why does the one-point-compactification not work? If its $D^2 backslash 0$, then the compactification is $D^2$ which maps by the identity to $D^2$.
$endgroup$
– user39598
Dec 18 '18 at 17:12
$begingroup$
This is true if you think of $D^2$ as the closed disk (which is a manifold with boundary), but not if you take it as the open disk, in which case the one-point compactification would be a pinched torus, not the closed disk.
$endgroup$
– Lukas Geyer
Dec 18 '18 at 17:17