Mixing
clarification about use of immersion in defining embedded submanifolds
$begingroup$
the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.Also, F(N) has a smooth structure naturally inherited from smooth structure on M and both of these stucture might be completely different as given in the example below:
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.)
I do understand that we want to investigate meaningful ways of endowing $pmb F(N)$ with a smooth submanifold structure. The word "submanifold" here suggests here that something about the smooth structure on $pmb F(N)$ needs to be compatible with the smooth structure on $pmb M$.But,as shown in example above, there are two smooth structures on $pmb F(N)$.
My question is:
Is there a role of taking $F$ as an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same?if not so,then why are immersions considered in definition??
Any input is welcome!
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
add a comment |
$begingroup$
the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.Also, F(N) has a smooth structure naturally inherited from smooth structure on M and both of these stucture might be completely different as given in the example below:
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.)
I do understand that we want to investigate meaningful ways of endowing $pmb F(N)$ with a smooth submanifold structure. The word "submanifold" here suggests here that something about the smooth structure on $pmb F(N)$ needs to be compatible with the smooth structure on $pmb M$.But,as shown in example above, there are two smooth structures on $pmb F(N)$.
My question is:
Is there a role of taking $F$ as an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same?if not so,then why are immersions considered in definition??
Any input is welcome!
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
$begingroup$
Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
$endgroup$
– Chris Huang
Jan 11 at 14:22
$begingroup$
@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
$endgroup$
– Abhishek Shrivastava
Jan 11 at 14:38
1
$begingroup$
Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
$endgroup$
– Moishe Cohen
Jan 12 at 16:45
add a comment |
$begingroup$
the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.Also, F(N) has a smooth structure naturally inherited from smooth structure on M and both of these stucture might be completely different as given in the example below:
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.)
I do understand that we want to investigate meaningful ways of endowing $pmb F(N)$ with a smooth submanifold structure. The word "submanifold" here suggests here that something about the smooth structure on $pmb F(N)$ needs to be compatible with the smooth structure on $pmb M$.But,as shown in example above, there are two smooth structures on $pmb F(N)$.
My question is:
Is there a role of taking $F$ as an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same?if not so,then why are immersions considered in definition??
Any input is welcome!
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.Also, F(N) has a smooth structure naturally inherited from smooth structure on M and both of these stucture might be completely different as given in the example below:
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.)
I do understand that we want to investigate meaningful ways of endowing $pmb F(N)$ with a smooth submanifold structure. The word "submanifold" here suggests here that something about the smooth structure on $pmb F(N)$ needs to be compatible with the smooth structure on $pmb M$.But,as shown in example above, there are two smooth structures on $pmb F(N)$.
My question is:
Is there a role of taking $F$ as an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same?if not so,then why are immersions considered in definition??
Any input is welcome!
geometry differential-geometry manifolds differential-topology smooth-manifolds
geometry differential-geometry manifolds differential-topology smooth-manifolds
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 11 at 14:01
Abhishek ShrivastavaAbhishek Shrivastava
414413
414413
$begingroup$
Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
$endgroup$
– Chris Huang
Jan 11 at 14:22
$begingroup$
@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
$endgroup$
– Abhishek Shrivastava
Jan 11 at 14:38
1
$begingroup$
Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
$endgroup$
– Moishe Cohen
Jan 12 at 16:45
add a comment |
$begingroup$
Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
$endgroup$
– Chris Huang
Jan 11 at 14:22
$begingroup$
@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
$endgroup$
– Abhishek Shrivastava
Jan 11 at 14:38
1
$begingroup$
Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
$endgroup$
– Moishe Cohen
Jan 12 at 16:45
$begingroup$
Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
$endgroup$
– Chris Huang
Jan 11 at 14:22
$begingroup$
Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
$endgroup$
– Chris Huang
Jan 11 at 14:22
$begingroup$
@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
$endgroup$
– Abhishek Shrivastava
Jan 11 at 14:38
$begingroup$
@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
$endgroup$
– Abhishek Shrivastava
Jan 11 at 14:38
1
1
$begingroup$
Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
$endgroup$
– Moishe Cohen
Jan 12 at 16:45
$begingroup$
Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
$endgroup$
– Moishe Cohen
Jan 12 at 16:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The comment by @Moishe Cohen sums up the reasons pretty well, but let me expand on them a little bit.
The most important reason for assuming $F$ is an immersion is that if you don't, then $F(N)$ can be a thing that we definitely do not want to consider as an immersed smooth submanifold. For example, consider the map $Fcolon mathbb Rto mathbb R^2$ given by
$$
F(x) = (x^3, x^2).
$$
This is a topological embedding and a smooth map. Its image has a cusp at the origin:
The whole idea of a smooth submanifold is that locally, it's supposed to look like a smoothly deformed version of a linear subspace. This does not.
The second reason for insisting on an immersion is that even in the case where the image set happens to be a nice smooth submanifold, the smooth structure on $N$ is supposed to have something to do with that on $M$. Yes, you can consider the case where they don't match, such as the map $Gcolon mathbb Rto mathbb R^2$ given by $G(x) = (x^3,0)$. The image set is the $x$-axis, but the smooth structure determined by $G$ is not related to the one it inherits from $mathbb R^2$. In this case, if you want to think of the image as a "subobject" of $mathbb R^2$, the original smooth structure is completely irrelevant. Requiring that the map be an immersion ensures that the smooth structure on $N$ is closely related to that of $M$.
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
$endgroup$
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
add a comment |
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$begingroup$
The comment by @Moishe Cohen sums up the reasons pretty well, but let me expand on them a little bit.
The most important reason for assuming $F$ is an immersion is that if you don't, then $F(N)$ can be a thing that we definitely do not want to consider as an immersed smooth submanifold. For example, consider the map $Fcolon mathbb Rto mathbb R^2$ given by
$$
F(x) = (x^3, x^2).
$$
This is a topological embedding and a smooth map. Its image has a cusp at the origin:
The whole idea of a smooth submanifold is that locally, it's supposed to look like a smoothly deformed version of a linear subspace. This does not.
The second reason for insisting on an immersion is that even in the case where the image set happens to be a nice smooth submanifold, the smooth structure on $N$ is supposed to have something to do with that on $M$. Yes, you can consider the case where they don't match, such as the map $Gcolon mathbb Rto mathbb R^2$ given by $G(x) = (x^3,0)$. The image set is the $x$-axis, but the smooth structure determined by $G$ is not related to the one it inherits from $mathbb R^2$. In this case, if you want to think of the image as a "subobject" of $mathbb R^2$, the original smooth structure is completely irrelevant. Requiring that the map be an immersion ensures that the smooth structure on $N$ is closely related to that of $M$.
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
$endgroup$
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
add a comment |
$begingroup$
The comment by @Moishe Cohen sums up the reasons pretty well, but let me expand on them a little bit.
The most important reason for assuming $F$ is an immersion is that if you don't, then $F(N)$ can be a thing that we definitely do not want to consider as an immersed smooth submanifold. For example, consider the map $Fcolon mathbb Rto mathbb R^2$ given by
$$
F(x) = (x^3, x^2).
$$
This is a topological embedding and a smooth map. Its image has a cusp at the origin:
The whole idea of a smooth submanifold is that locally, it's supposed to look like a smoothly deformed version of a linear subspace. This does not.
The second reason for insisting on an immersion is that even in the case where the image set happens to be a nice smooth submanifold, the smooth structure on $N$ is supposed to have something to do with that on $M$. Yes, you can consider the case where they don't match, such as the map $Gcolon mathbb Rto mathbb R^2$ given by $G(x) = (x^3,0)$. The image set is the $x$-axis, but the smooth structure determined by $G$ is not related to the one it inherits from $mathbb R^2$. In this case, if you want to think of the image as a "subobject" of $mathbb R^2$, the original smooth structure is completely irrelevant. Requiring that the map be an immersion ensures that the smooth structure on $N$ is closely related to that of $M$.
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
$endgroup$
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
add a comment |
$begingroup$
The comment by @Moishe Cohen sums up the reasons pretty well, but let me expand on them a little bit.
The most important reason for assuming $F$ is an immersion is that if you don't, then $F(N)$ can be a thing that we definitely do not want to consider as an immersed smooth submanifold. For example, consider the map $Fcolon mathbb Rto mathbb R^2$ given by
$$
F(x) = (x^3, x^2).
$$
This is a topological embedding and a smooth map. Its image has a cusp at the origin:
The whole idea of a smooth submanifold is that locally, it's supposed to look like a smoothly deformed version of a linear subspace. This does not.
The second reason for insisting on an immersion is that even in the case where the image set happens to be a nice smooth submanifold, the smooth structure on $N$ is supposed to have something to do with that on $M$. Yes, you can consider the case where they don't match, such as the map $Gcolon mathbb Rto mathbb R^2$ given by $G(x) = (x^3,0)$. The image set is the $x$-axis, but the smooth structure determined by $G$ is not related to the one it inherits from $mathbb R^2$. In this case, if you want to think of the image as a "subobject" of $mathbb R^2$, the original smooth structure is completely irrelevant. Requiring that the map be an immersion ensures that the smooth structure on $N$ is closely related to that of $M$.
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
$endgroup$
The comment by @Moishe Cohen sums up the reasons pretty well, but let me expand on them a little bit.
The most important reason for assuming $F$ is an immersion is that if you don't, then $F(N)$ can be a thing that we definitely do not want to consider as an immersed smooth submanifold. For example, consider the map $Fcolon mathbb Rto mathbb R^2$ given by
$$
F(x) = (x^3, x^2).
$$
This is a topological embedding and a smooth map. Its image has a cusp at the origin:
The whole idea of a smooth submanifold is that locally, it's supposed to look like a smoothly deformed version of a linear subspace. This does not.
The second reason for insisting on an immersion is that even in the case where the image set happens to be a nice smooth submanifold, the smooth structure on $N$ is supposed to have something to do with that on $M$. Yes, you can consider the case where they don't match, such as the map $Gcolon mathbb Rto mathbb R^2$ given by $G(x) = (x^3,0)$. The image set is the $x$-axis, but the smooth structure determined by $G$ is not related to the one it inherits from $mathbb R^2$. In this case, if you want to think of the image as a "subobject" of $mathbb R^2$, the original smooth structure is completely irrelevant. Requiring that the map be an immersion ensures that the smooth structure on $N$ is closely related to that of $M$.
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
answered Jan 12 at 19:20
Jack LeeJack Lee
27.3k54666
27.3k54666
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
add a comment |
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:thanks for your answer sir...
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:35
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
:moreover, can we also prove that smooth structure inherited on F(N) from M and smooth structure derived by taking topological embedding are equivalent iff f is an immersion
$endgroup$
– Abhishek Shrivastava
Jan 12 at 19:38
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
$begingroup$
@AbhishekShrivastava: You're welcome. Yes -- IF you assume that $F(N)$ is an embedded submanifold of $M$ and $F$ is a smooth topological embedding, then those two smooth structures are the same iff $F$ is a diffeomorphism onto its image. And once you know $F$ is a smooth topological embedding, it's a diffeo onto its image if and only if it's an immersion.
$endgroup$
– Jack Lee
Jan 12 at 20:24
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Basically the immersion condition makes sure that the image does not have ‘cusps’ and so is a smooth submanifold, rather than a topological submanifold that is singular.
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– Chris Huang
Jan 11 at 14:22
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@Chris Huang:thanks for your view...i do get this,but how does immersion ensures this and how do the both smooth structure, (one defined by diffeomorphism through topological embedding(by push forward) and other one inherited by smooth structure in which F(N) sits)become same merely by consideration of immersions
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– Abhishek Shrivastava
Jan 11 at 14:38
1
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Yes, the answer to your question is positive: The immersion condition ensures that (1) $F(N)$ is a smooth submanifold of $M$ and (2) the two smooth structures on $F(N)$ are the same. As a more extreme example, consider manifolds $N, M$ of the same dimension.
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– Moishe Cohen
Jan 12 at 16:45