A connected embedded submanifold, which is contained in an immersed submanifold, is connected in this...
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I'm currently studying foliation theory, and to solve a problem I need the following to be true, but I can't prove nor disprove, I just have a feeling it may be false in general: if I have a smooth manifold $M,$ and a connected embedded submanifold $P$ of $M,$ which is contained in an immersed submanifold $F$ of $M,$ is $P$ also connected in the immersed submanifold topology of $F$?
At first I thought this was obviously true, but giving it a little more thinking, either I'm making some mistake or, if this is true, it is certainly not obvious. Here is what I did: $F$ may have a finer topology than its subspace topology. If $F$ was an embedded manifold, then the subspace topology of $P$ in $M$ would coincide with its the subspace topology as a subset of $F,$and then it would be connected in $F$. But with $F$ being immersed, $P$ with a topology induced by the immersed topology of $F$ may have an open set which messes with its connectedness. Any ideas, or did I made some mistake? Thanks.
general-topology manifolds connectedness
asked Dec 24 '18 at 6:31
VicVic
4217
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add a comment |
$begingroup$
I'm currently studying foliation theory, and to solve a problem I need the following to be true, but I can't prove nor disprove, I just have a feeling it may be false in general: if I have a smooth manifold $M,$ and a connected embedded submanifold $P$ of $M,$ which is contained in an immersed submanifold $F$ of $M,$ is $P$ also connected in the immersed submanifold topology of $F$?
At first I thought this was obviously true, but giving it a little more thinking, either I'm making some mistake or, if this is true, it is certainly not obvious. Here is what I did: $F$ may have a finer topology than its subspace topology. If $F$ was an embedded manifold, then the subspace topology of $P$ in $M$ would coincide with its the subspace topology as a subset of $F,$and then it would be connected in $F$. But with $F$ being immersed, $P$ with a topology induced by the immersed topology of $F$ may have an open set which messes with its connectedness. Any ideas, or did I made some mistake? Thanks.
general-topology manifolds connectedness
asked Dec 24 '18 at 6:31
VicVic
4217
$endgroup$
add a comment |
$begingroup$
I'm currently studying foliation theory, and to solve a problem I need the following to be true, but I can't prove nor disprove, I just have a feeling it may be false in general: if I have a smooth manifold $M,$ and a connected embedded submanifold $P$ of $M,$ which is contained in an immersed submanifold $F$ of $M,$ is $P$ also connected in the immersed submanifold topology of $F$?
At first I thought this was obviously true, but giving it a little more thinking, either I'm making some mistake or, if this is true, it is certainly not obvious. Here is what I did: $F$ may have a finer topology than its subspace topology. If $F$ was an embedded manifold, then the subspace topology of $P$ in $M$ would coincide with its the subspace topology as a subset of $F,$and then it would be connected in $F$. But with $F$ being immersed, $P$ with a topology induced by the immersed topology of $F$ may have an open set which messes with its connectedness. Any ideas, or did I made some mistake? Thanks.
general-topology manifolds connectedness
asked Dec 24 '18 at 6:31
VicVic
4217
$endgroup$
I'm currently studying foliation theory, and to solve a problem I need the following to be true, but I can't prove nor disprove, I just have a feeling it may be false in general: if I have a smooth manifold $M,$ and a connected embedded submanifold $P$ of $M,$ which is contained in an immersed submanifold $F$ of $M,$ is $P$ also connected in the immersed submanifold topology of $F$?
At first I thought this was obviously true, but giving it a little more thinking, either I'm making some mistake or, if this is true, it is certainly not obvious. Here is what I did: $F$ may have a finer topology than its subspace topology. If $F$ was an embedded manifold, then the subspace topology of $P$ in $M$ would coincide with its the subspace topology as a subset of $F,$and then it would be connected in $F$. But with $F$ being immersed, $P$ with a topology induced by the immersed topology of $F$ may have an open set which messes with its connectedness. Any ideas, or did I made some mistake? Thanks.
general-topology manifolds connectedness
general-topology manifolds connectedness
asked Dec 24 '18 at 6:31
VicVic
4217
asked Dec 24 '18 at 6:31
VicVic
4217
edited Dec 24 '18 at 6:48
Vic
asked Dec 24 '18 at 6:31
VicVic
4217
asked Dec 24 '18 at 6:31
VicVic
4217
asked Dec 24 '18 at 6:31
VicVic
4217
4217
add a comment |
add a comment |
1 Answer
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This is in fact false in general. Let $M=mathbb{R}^2$ and let $F$ be an open interval immersed in $M$ in the shape of the numeral $$mathsf{8},$$ starting from the middle point and going around the top loop counterclockwise and then around the bottom loop clockwise. Let $P$ be a subset of the $mathsf{8}$ which starts at the bottom left and then goes diagonally to the top right. Then $P$ is of course a connected embedded submanifold of $M$. However, $P$ is not connected in the topology of $F$, since it consists of two separate intervals at the start and end of $F$ together with one point (the middle point) which is halfway between them.
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
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$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is in fact false in general. Let $M=mathbb{R}^2$ and let $F$ be an open interval immersed in $M$ in the shape of the numeral $$mathsf{8},$$ starting from the middle point and going around the top loop counterclockwise and then around the bottom loop clockwise. Let $P$ be a subset of the $mathsf{8}$ which starts at the bottom left and then goes diagonally to the top right. Then $P$ is of course a connected embedded submanifold of $M$. However, $P$ is not connected in the topology of $F$, since it consists of two separate intervals at the start and end of $F$ together with one point (the middle point) which is halfway between them.
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
add a comment |
$begingroup$
This is in fact false in general. Let $M=mathbb{R}^2$ and let $F$ be an open interval immersed in $M$ in the shape of the numeral $$mathsf{8},$$ starting from the middle point and going around the top loop counterclockwise and then around the bottom loop clockwise. Let $P$ be a subset of the $mathsf{8}$ which starts at the bottom left and then goes diagonally to the top right. Then $P$ is of course a connected embedded submanifold of $M$. However, $P$ is not connected in the topology of $F$, since it consists of two separate intervals at the start and end of $F$ together with one point (the middle point) which is halfway between them.
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
add a comment |
$begingroup$
This is in fact false in general. Let $M=mathbb{R}^2$ and let $F$ be an open interval immersed in $M$ in the shape of the numeral $$mathsf{8},$$ starting from the middle point and going around the top loop counterclockwise and then around the bottom loop clockwise. Let $P$ be a subset of the $mathsf{8}$ which starts at the bottom left and then goes diagonally to the top right. Then $P$ is of course a connected embedded submanifold of $M$. However, $P$ is not connected in the topology of $F$, since it consists of two separate intervals at the start and end of $F$ together with one point (the middle point) which is halfway between them.
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
$endgroup$
This is in fact false in general. Let $M=mathbb{R}^2$ and let $F$ be an open interval immersed in $M$ in the shape of the numeral $$mathsf{8},$$ starting from the middle point and going around the top loop counterclockwise and then around the bottom loop clockwise. Let $P$ be a subset of the $mathsf{8}$ which starts at the bottom left and then goes diagonally to the top right. Then $P$ is of course a connected embedded submanifold of $M$. However, $P$ is not connected in the topology of $F$, since it consists of two separate intervals at the start and end of $F$ together with one point (the middle point) which is halfway between them.
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
answered Dec 24 '18 at 7:05
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
add a comment |
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
$begingroup$
And the immersed 8 figure saves the day again. Thanks!
$endgroup$
– Vic
Dec 24 '18 at 19:45
add a comment |
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