“Lift” of an immersion

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Let $f : Sigma^k to M^n$ be an immersion between differentiable manifolds and let $pi : tildeM to M$ be a finite-to-one covering map. Let $tildeSigma = pi^-1(f(Sigma))$. Is it true that $tildeSigma$ is immersed somehow into $tildeM$?

If $f$ is an embedding, I know how to prove that $tildeSigma$ is embedded into $tildeM$, since we can use the fact that $pi$ is transversal to $f(Sigma) cong Sigma$ (as $pi$ is a submersion) to conclude that $pi^-1(f(Sigma))$ is a submanifold of $tildeM$. What about the general case?

differential-geometry manifolds smooth-manifolds covering-spaces submanifold

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edited Jul 25 at 18:21

asked Jul 25 at 18:04

Eduardo Longa

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up vote
2
down vote

favorite

1

Let $f : Sigma^k to M^n$ be an immersion between differentiable manifolds and let $pi : tildeM to M$ be a finite-to-one covering map. Let $tildeSigma = pi^-1(f(Sigma))$. Is it true that $tildeSigma$ is immersed somehow into $tildeM$?

If $f$ is an embedding, I know how to prove that $tildeSigma$ is embedded into $tildeM$, since we can use the fact that $pi$ is transversal to $f(Sigma) cong Sigma$ (as $pi$ is a submersion) to conclude that $pi^-1(f(Sigma))$ is a submanifold of $tildeM$. What about the general case?

differential-geometry manifolds smooth-manifolds covering-spaces submanifold

share|cite|improve this question

edited Jul 25 at 18:21

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Let $f : Sigma^k to M^n$ be an immersion between differentiable manifolds and let $pi : tildeM to M$ be a finite-to-one covering map. Let $tildeSigma = pi^-1(f(Sigma))$. Is it true that $tildeSigma$ is immersed somehow into $tildeM$?

If $f$ is an embedding, I know how to prove that $tildeSigma$ is embedded into $tildeM$, since we can use the fact that $pi$ is transversal to $f(Sigma) cong Sigma$ (as $pi$ is a submersion) to conclude that $pi^-1(f(Sigma))$ is a submanifold of $tildeM$. What about the general case?

differential-geometry manifolds smooth-manifolds covering-spaces submanifold

share|cite|improve this question

edited Jul 25 at 18:21

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

Let $f : Sigma^k to M^n$ be an immersion between differentiable manifolds and let $pi : tildeM to M$ be a finite-to-one covering map. Let $tildeSigma = pi^-1(f(Sigma))$. Is it true that $tildeSigma$ is immersed somehow into $tildeM$?

If $f$ is an embedding, I know how to prove that $tildeSigma$ is embedded into $tildeM$, since we can use the fact that $pi$ is transversal to $f(Sigma) cong Sigma$ (as $pi$ is a submersion) to conclude that $pi^-1(f(Sigma))$ is a submanifold of $tildeM$. What about the general case?

differential-geometry manifolds smooth-manifolds covering-spaces submanifold

share|cite|improve this question

edited Jul 25 at 18:21

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 18:21

edited Jul 25 at 18:21

edited Jul 25 at 18:21

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

asked Jul 25 at 18:04

Eduardo Longa

1,5062518

1,5062518

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

You can take the pullback of $tildeM$ along $f$, as depicted in the following diagram: $$beginarraycccf^*tildeM&longrightarrow&tildeM\downarrow&quad&downarrow\Sigma&xrightarrowf&Mendarray.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*tildeMtotildeM$ is an immersion with image $pi^-1(f(Sigma)).$

Edit: We explain why $f^*tildeMtotildeM$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*tildeMtoSigmato M$$ is an immersion. As the diagram commutes, the composition $$f^*tildeMtotildeMto M$$ is also an immersion, and consequently, so is the left-hand arrow in it.

share|cite|improve this answer

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
  – Eduardo Longa
  Jul 25 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoLonga Please check the edit.
  – Amitai Yuval
  Jul 26 at 7:53
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

You can take the pullback of $tildeM$ along $f$, as depicted in the following diagram: $$beginarraycccf^*tildeM&longrightarrow&tildeM\downarrow&quad&downarrow\Sigma&xrightarrowf&Mendarray.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*tildeMtotildeM$ is an immersion with image $pi^-1(f(Sigma)).$

Edit: We explain why $f^*tildeMtotildeM$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*tildeMtoSigmato M$$ is an immersion. As the diagram commutes, the composition $$f^*tildeMtotildeMto M$$ is also an immersion, and consequently, so is the left-hand arrow in it.

share|cite|improve this answer

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
  – Eduardo Longa
  Jul 25 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoLonga Please check the edit.
  – Amitai Yuval
  Jul 26 at 7:53
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

You can take the pullback of $tildeM$ along $f$, as depicted in the following diagram: $$beginarraycccf^*tildeM&longrightarrow&tildeM\downarrow&quad&downarrow\Sigma&xrightarrowf&Mendarray.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*tildeMtotildeM$ is an immersion with image $pi^-1(f(Sigma)).$

Edit: We explain why $f^*tildeMtotildeM$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*tildeMtoSigmato M$$ is an immersion. As the diagram commutes, the composition $$f^*tildeMtotildeMto M$$ is also an immersion, and consequently, so is the left-hand arrow in it.

share|cite|improve this answer

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
  – Eduardo Longa
  Jul 25 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoLonga Please check the edit.
  – Amitai Yuval
  Jul 26 at 7:53
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

You can take the pullback of $tildeM$ along $f$, as depicted in the following diagram: $$beginarraycccf^*tildeM&longrightarrow&tildeM\downarrow&quad&downarrow\Sigma&xrightarrowf&Mendarray.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*tildeMtotildeM$ is an immersion with image $pi^-1(f(Sigma)).$

Edit: We explain why $f^*tildeMtotildeM$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*tildeMtoSigmato M$$ is an immersion. As the diagram commutes, the composition $$f^*tildeMtotildeMto M$$ is also an immersion, and consequently, so is the left-hand arrow in it.

share|cite|improve this answer

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

You can take the pullback of $tildeM$ along $f$, as depicted in the following diagram: $$beginarraycccf^*tildeM&longrightarrow&tildeM\downarrow&quad&downarrow\Sigma&xrightarrowf&Mendarray.$$ The pullback can be described in a few different ways. In any case, as the above diagram and the corresponding diagram of tangent bundles commute, it follows that $f^*tildeMtotildeM$ is an immersion with image $pi^-1(f(Sigma)).$

Edit: We explain why $f^*tildeMtotildeM$ is an immersion. Both downwards-pointing arrows in the diagram are covering maps and, in particular, local diffeomorphisms. Hence, the composition $$f^*tildeMtoSigmato M$$ is an immersion. As the diagram commutes, the composition $$f^*tildeMtotildeMto M$$ is also an immersion, and consequently, so is the left-hand arrow in it.

share|cite|improve this answer

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 7:52

edited Jul 26 at 7:52

edited Jul 26 at 7:52

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

answered Jul 25 at 20:30

Amitai Yuval

14.4k11026

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
  – Eduardo Longa
  Jul 25 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoLonga Please check the edit.
  – Amitai Yuval
  Jul 26 at 7:53
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
  – Eduardo Longa
  Jul 25 at 20:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoLonga Please check the edit.
  – Amitai Yuval
  Jul 26 at 7:53
  

  
  
  
  

Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
– Eduardo Longa
Jul 25 at 20:59

Could you provide more details? Why is $f^* tildeM to tildeM$ an immersion?
– Eduardo Longa
Jul 25 at 20:59

@EduardoLonga Please check the edit.
– Amitai Yuval
Jul 26 at 7:53

@EduardoLonga Please check the edit.
– Amitai Yuval
Jul 26 at 7:53

add a comment | 

 

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