Is the pullback of a submersion map injective or surjective? [closed]
A very basic question.
Let $f:Mlongrightarrow N$ be a smooth submersion map, we know that its pushforward $f_*:TMlongrightarrow TN$ is surjective. Then what about its pullback $f^*:T^*Nlongrightarrow T^*M$? Is it injective or surjective?
linear-algebra
asked Nov 25 at 8:49
user605302
234
closed as off-topic by Travis, Saad, Davide Giraudo, Paul Frost, amWhy Nov 25 at 20:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Saad, Davide Giraudo, Paul Frost, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
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A very basic question.
Let $f:Mlongrightarrow N$ be a smooth submersion map, we know that its pushforward $f_*:TMlongrightarrow TN$ is surjective. Then what about its pullback $f^*:T^*Nlongrightarrow T^*M$? Is it injective or surjective?
linear-algebra
asked Nov 25 at 8:49
user605302
234
closed as off-topic by Travis, Saad, Davide Giraudo, Paul Frost, amWhy Nov 25 at 20:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Saad, Davide Giraudo, Paul Frost, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36
add a comment |
A very basic question.
Let $f:Mlongrightarrow N$ be a smooth submersion map, we know that its pushforward $f_*:TMlongrightarrow TN$ is surjective. Then what about its pullback $f^*:T^*Nlongrightarrow T^*M$? Is it injective or surjective?
linear-algebra
asked Nov 25 at 8:49
user605302
234
A very basic question.
Let $f:Mlongrightarrow N$ be a smooth submersion map, we know that its pushforward $f_*:TMlongrightarrow TN$ is surjective. Then what about its pullback $f^*:T^*Nlongrightarrow T^*M$? Is it injective or surjective?
linear-algebra
linear-algebra
asked Nov 25 at 8:49
user605302
234
asked Nov 25 at 8:49
user605302
234
edited Nov 25 at 9:36
asked Nov 25 at 8:49
user605302
234
asked Nov 25 at 8:49
user605302
234
asked Nov 25 at 8:49
user605302
234
234
closed as off-topic by Travis, Saad, Davide Giraudo, Paul Frost, amWhy Nov 25 at 20:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Saad, Davide Giraudo, Paul Frost, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Travis, Saad, Davide Giraudo, Paul Frost, amWhy Nov 25 at 20:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, Saad, Davide Giraudo, Paul Frost, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
1
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36
add a comment |
1
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36
1
1
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36
add a comment |
1 Answer
1
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oldest
votes
It is injective Infact if you consider
$w,theta in T^*N$ such that $f^*(w)=f^*(theta)$ then for every $vin T N$ you have that there exists $ain T M$ such that $v=f_*(a)$ and so
$w(v)=w(f_*(a))=wcirc f_*(a)=f^*(w)(a)=$
$f^*(theta)(a)=theta(f_*(a))=theta(v) $
Then
$w(v)=theta(v)$ so $w=theta$
There is a problem for the surjectivity because if you consider
$win T^*M $ you want built $theta: TNto mathbb{R}$ such that $f^*(theta)=w$
The idea could be to define for all $vin TN$
$theta(v):=w(a)$ where $a$ is an element of $TM$ such that $f_*(a)=v$
The problem is that there could be two different $a,bin TM$ such that $f_*(a)=f_*(b)=v$ but in general there is the possibility that $w(a)neq w(b)$ so $theta$ is not well defined.
You can observe that all proofs are not dipend by the contest of the differential geometry, Infact these are general results of linear geometry.
answered Nov 25 at 9:06
Federico Fallucca
1,81318
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is injective Infact if you consider
$w,theta in T^*N$ such that $f^*(w)=f^*(theta)$ then for every $vin T N$ you have that there exists $ain T M$ such that $v=f_*(a)$ and so
$w(v)=w(f_*(a))=wcirc f_*(a)=f^*(w)(a)=$
$f^*(theta)(a)=theta(f_*(a))=theta(v) $
Then
$w(v)=theta(v)$ so $w=theta$
There is a problem for the surjectivity because if you consider
$win T^*M $ you want built $theta: TNto mathbb{R}$ such that $f^*(theta)=w$
The idea could be to define for all $vin TN$
$theta(v):=w(a)$ where $a$ is an element of $TM$ such that $f_*(a)=v$
The problem is that there could be two different $a,bin TM$ such that $f_*(a)=f_*(b)=v$ but in general there is the possibility that $w(a)neq w(b)$ so $theta$ is not well defined.
You can observe that all proofs are not dipend by the contest of the differential geometry, Infact these are general results of linear geometry.
answered Nov 25 at 9:06
Federico Fallucca
1,81318
add a comment |
It is injective Infact if you consider
$w,theta in T^*N$ such that $f^*(w)=f^*(theta)$ then for every $vin T N$ you have that there exists $ain T M$ such that $v=f_*(a)$ and so
$w(v)=w(f_*(a))=wcirc f_*(a)=f^*(w)(a)=$
$f^*(theta)(a)=theta(f_*(a))=theta(v) $
Then
$w(v)=theta(v)$ so $w=theta$
There is a problem for the surjectivity because if you consider
$win T^*M $ you want built $theta: TNto mathbb{R}$ such that $f^*(theta)=w$
The idea could be to define for all $vin TN$
$theta(v):=w(a)$ where $a$ is an element of $TM$ such that $f_*(a)=v$
The problem is that there could be two different $a,bin TM$ such that $f_*(a)=f_*(b)=v$ but in general there is the possibility that $w(a)neq w(b)$ so $theta$ is not well defined.
You can observe that all proofs are not dipend by the contest of the differential geometry, Infact these are general results of linear geometry.
answered Nov 25 at 9:06
Federico Fallucca
1,81318
add a comment |
It is injective Infact if you consider
$w,theta in T^*N$ such that $f^*(w)=f^*(theta)$ then for every $vin T N$ you have that there exists $ain T M$ such that $v=f_*(a)$ and so
$w(v)=w(f_*(a))=wcirc f_*(a)=f^*(w)(a)=$
$f^*(theta)(a)=theta(f_*(a))=theta(v) $
Then
$w(v)=theta(v)$ so $w=theta$
There is a problem for the surjectivity because if you consider
$win T^*M $ you want built $theta: TNto mathbb{R}$ such that $f^*(theta)=w$
The idea could be to define for all $vin TN$
$theta(v):=w(a)$ where $a$ is an element of $TM$ such that $f_*(a)=v$
The problem is that there could be two different $a,bin TM$ such that $f_*(a)=f_*(b)=v$ but in general there is the possibility that $w(a)neq w(b)$ so $theta$ is not well defined.
You can observe that all proofs are not dipend by the contest of the differential geometry, Infact these are general results of linear geometry.
answered Nov 25 at 9:06
Federico Fallucca
1,81318
It is injective Infact if you consider
$w,theta in T^*N$ such that $f^*(w)=f^*(theta)$ then for every $vin T N$ you have that there exists $ain T M$ such that $v=f_*(a)$ and so
$w(v)=w(f_*(a))=wcirc f_*(a)=f^*(w)(a)=$
$f^*(theta)(a)=theta(f_*(a))=theta(v) $
Then
$w(v)=theta(v)$ so $w=theta$
There is a problem for the surjectivity because if you consider
$win T^*M $ you want built $theta: TNto mathbb{R}$ such that $f^*(theta)=w$
The idea could be to define for all $vin TN$
$theta(v):=w(a)$ where $a$ is an element of $TM$ such that $f_*(a)=v$
The problem is that there could be two different $a,bin TM$ such that $f_*(a)=f_*(b)=v$ but in general there is the possibility that $w(a)neq w(b)$ so $theta$ is not well defined.
You can observe that all proofs are not dipend by the contest of the differential geometry, Infact these are general results of linear geometry.
answered Nov 25 at 9:06
Federico Fallucca
1,81318
edited Nov 25 at 9:16
answered Nov 25 at 9:06
Federico Fallucca
1,81318
answered Nov 25 at 9:06
Federico Fallucca
1,81318
answered Nov 25 at 9:06
Federico Fallucca
1,81318
1,81318
add a comment |
add a comment |
1
The way you’ve written things, $f_*$ is not surjective unless you assume that $f$ is.
– Matthew Kvalheim
Nov 25 at 9:36