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Prove that if $f$ is a diffeomorphism than its differential $D_f$ is an isomorphism

If $T:V to W$ is such that both $ker(T)$ and $operatornameIm(T)$ are finite-dimensional, then $V$ is finite-dimensionalProve that a linear map $A$ does not have an inverse map if $A circ A$ = O where $A$ is a linear transformation from X onto itselfHow do I prove that if T is surjective, then T is an isomorphism.Showing existence of Linear transformation such that $T(v)neq0$, for any non zero vector vIf $T:Vto W$ is an injective linear transformation and $T^*:Wto V$ is its adjoint, is $T^*T:Vto V$ necessarily an isomorphismAre the following two statements about linear transforms true?$mathcal L(V_1 times cdots times V_m,W) approx mathcal L(V_1,W) times cdots times mathcal L(V_m,W)$Proving that the linear mapping $f: P(x,Bbb R) to P(x,Bbb R)$, which maps the polynomial $p(x)$ to the polynomial $p(x-1)$, is the isomorphismIsomorphism for finite dimensional Vector Spaces of same dimension over field FWhat is the inverse of this linear map?

0

$begingroup$

I want to prove that if $f : U to V$ , when $U,V subset R^n$, is a diffeomorphism than its differential $D_f$ is an isomorphism.

Since $D_f$ is a linear transformation, isomorphism means that it's both injective and surjective.
Moreover, $D_f$ is between two vector spaces with the same dimension (both are $R^n*n$) then we can use the dimension theorem for vector spaces. So we really need to prove only that $D_f$ is injective (or surjective).

However, I don't really see how to do it. I know that $f$ is $C^1$ (continuously differential) and also $f^-1$ is $C^1$.

Maybe, because the inverse of $D_f$ is $D_f^-1$. So if $D_f$ is not injective then it doesn't have an inverse - contradiction. I'm not really sure if I'm right here.

Help would be appreciated.

multivariable-calculus linear-transformations vector-space-isomorphism diffeomorphism

share|cite|improve this question

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The differential satisfies $D_fcirc g=D_fcirc D_g$ and the differential of identity (as a diffeomorphism) is again the identity (as a linear map). These two properties are also know as "functoriality". Can you see how to apply it to obtain that $D_f$ has to be invertible? Also note that "isomorphism" means "an invertible morphism" by definition. In case of linear maps this coincides with "being bijective".
  $endgroup$
  – freakish
  Mar 21 at 23:36
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I want to prove that if $f : U to V$ , when $U,V subset R^n$, is a diffeomorphism than its differential $D_f$ is an isomorphism.

Since $D_f$ is a linear transformation, isomorphism means that it's both injective and surjective.
Moreover, $D_f$ is between two vector spaces with the same dimension (both are $R^n*n$) then we can use the dimension theorem for vector spaces. So we really need to prove only that $D_f$ is injective (or surjective).

However, I don't really see how to do it. I know that $f$ is $C^1$ (continuously differential) and also $f^-1$ is $C^1$.

Maybe, because the inverse of $D_f$ is $D_f^-1$. So if $D_f$ is not injective then it doesn't have an inverse - contradiction. I'm not really sure if I'm right here.

Help would be appreciated.

multivariable-calculus linear-transformations vector-space-isomorphism diffeomorphism

share|cite|improve this question

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The differential satisfies $D_fcirc g=D_fcirc D_g$ and the differential of identity (as a diffeomorphism) is again the identity (as a linear map). These two properties are also know as "functoriality". Can you see how to apply it to obtain that $D_f$ has to be invertible? Also note that "isomorphism" means "an invertible morphism" by definition. In case of linear maps this coincides with "being bijective".
  $endgroup$
  – freakish
  Mar 21 at 23:36
  
  

  
  
  
  

add a comment | 

$begingroup$

I want to prove that if $f : U to V$ , when $U,V subset R^n$, is a diffeomorphism than its differential $D_f$ is an isomorphism.

Since $D_f$ is a linear transformation, isomorphism means that it's both injective and surjective.
Moreover, $D_f$ is between two vector spaces with the same dimension (both are $R^n*n$) then we can use the dimension theorem for vector spaces. So we really need to prove only that $D_f$ is injective (or surjective).

However, I don't really see how to do it. I know that $f$ is $C^1$ (continuously differential) and also $f^-1$ is $C^1$.

Maybe, because the inverse of $D_f$ is $D_f^-1$. So if $D_f$ is not injective then it doesn't have an inverse - contradiction. I'm not really sure if I'm right here.

Help would be appreciated.

multivariable-calculus linear-transformations vector-space-isomorphism diffeomorphism

share|cite|improve this question

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

$endgroup$

share|cite|improve this question

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

share|cite|improve this question

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

edited Mar 21 at 23:55

José Carlos Santos

173k23133241

asked Mar 21 at 23:32

Gabi GGabi G

500110

asked Mar 21 at 23:32

Gabi GGabi G

500110

2 Answers
2

active

oldest

votes

2

$begingroup$

Since $f^-1circ f=operatornameId$, $D_f^-1circ D_f=D_f^-1circ f=operatornameId$. But $D_f$ and $D_f^-1$ are linear maps from $mathbbR^n$ into itself. So, each ne is the inverse of the other one.

share|cite|improve this answer

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so it's pretty much what I wrote, right?
  $endgroup$
  – Gabi G
  Mar 21 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. You didn't seem sure about it and I don't see where your doubt lies.
  $endgroup$
  – José Carlos Santos
  Mar 21 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually though about this while writing the question
  $endgroup$
  – Gabi G
  Mar 21 at 23:48
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Well, if $f$ is a $mathscrC^r$ diffeomorphism, then it has $mathscrC^r$ inverse called $g$. We know that the differential $D$ has the property $D(fcirc g)=Dfcirc Dg$ by the chain rule. So, since $fcirc g= mathbb1_V$ and $gcirc f=mathbb1_U$, we get that $Df$ and $Dg$ are mutually inverse, hence isomorphisms.

share|cite|improve this answer

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

add a comment | 

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2

$begingroup$

Since $f^-1circ f=operatornameId$, $D_f^-1circ D_f=D_f^-1circ f=operatornameId$. But $D_f$ and $D_f^-1$ are linear maps from $mathbbR^n$ into itself. So, each ne is the inverse of the other one.

share|cite|improve this answer

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so it's pretty much what I wrote, right?
  $endgroup$
  – Gabi G
  Mar 21 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. You didn't seem sure about it and I don't see where your doubt lies.
  $endgroup$
  – José Carlos Santos
  Mar 21 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually though about this while writing the question
  $endgroup$
  – Gabi G
  Mar 21 at 23:48
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Since $f^-1circ f=operatornameId$, $D_f^-1circ D_f=D_f^-1circ f=operatornameId$. But $D_f$ and $D_f^-1$ are linear maps from $mathbbR^n$ into itself. So, each ne is the inverse of the other one.

share|cite|improve this answer

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so it's pretty much what I wrote, right?
  $endgroup$
  – Gabi G
  Mar 21 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. You didn't seem sure about it and I don't see where your doubt lies.
  $endgroup$
  – José Carlos Santos
  Mar 21 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I actually though about this while writing the question
  $endgroup$
  – Gabi G
  Mar 21 at 23:48
  
  

  
  
  
  

add a comment | 

$begingroup$

Since $f^-1circ f=operatornameId$, $D_f^-1circ D_f=D_f^-1circ f=operatornameId$. But $D_f$ and $D_f^-1$ are linear maps from $mathbbR^n$ into itself. So, each ne is the inverse of the other one.

share|cite|improve this answer

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

answered Mar 21 at 23:36

José Carlos SantosJosé Carlos Santos

173k23133241

1

$begingroup$

Well, if $f$ is a $mathscrC^r$ diffeomorphism, then it has $mathscrC^r$ inverse called $g$. We know that the differential $D$ has the property $D(fcirc g)=Dfcirc Dg$ by the chain rule. So, since $fcirc g= mathbb1_V$ and $gcirc f=mathbb1_U$, we get that $Df$ and $Dg$ are mutually inverse, hence isomorphisms.

share|cite|improve this answer

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

add a comment | 

1

$begingroup$

Well, if $f$ is a $mathscrC^r$ diffeomorphism, then it has $mathscrC^r$ inverse called $g$. We know that the differential $D$ has the property $D(fcirc g)=Dfcirc Dg$ by the chain rule. So, since $fcirc g= mathbb1_V$ and $gcirc f=mathbb1_U$, we get that $Df$ and $Dg$ are mutually inverse, hence isomorphisms.

share|cite|improve this answer

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

add a comment | 

$begingroup$

Well, if $f$ is a $mathscrC^r$ diffeomorphism, then it has $mathscrC^r$ inverse called $g$. We know that the differential $D$ has the property $D(fcirc g)=Dfcirc Dg$ by the chain rule. So, since $fcirc g= mathbb1_V$ and $gcirc f=mathbb1_U$, we get that $Df$ and $Dg$ are mutually inverse, hence isomorphisms.

share|cite|improve this answer

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

share|cite|improve this answer

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

answered Mar 21 at 23:38

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741