About the Jacobian determinant and the bijection

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Discussion Overview

The discussion revolves around the inverse function theorem and its conditions for local invertibility, specifically the relationship between the Jacobian determinant and the bijectiveness of the derivative of a function. Participants explore different formulations and interpretations of these conditions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that some texts state local invertibility requires the function to be C1 and the Jacobian determinant to be non-zero, while others suggest that bijectiveness of the derivative is sufficient.
  • Another participant asserts that the bijectiveness of the linear map F'(a) is equivalent to the determinant of F'(a) being non-zero, referencing linear algebra principles.
  • A participant inquires about the name of the proposition related to these conditions, suggesting that it may relate to the invertibility of matrices and vector space isomorphisms.
  • One participant mentions the "implicit function theorem" as a general theorem that states if the Jacobian is non-zero at a point, then one can solve for one variable as a function of the others in a neighborhood around that point.

Areas of Agreement / Disagreement

Participants generally agree on the equivalence of the bijectiveness of the derivative and the non-zero determinant, but there is uncertainty regarding the specific naming of the propositions and the broader implications of these conditions.

Contextual Notes

There is a lack of consensus on the specific naming of the propositions discussed, and the implications of the implicit function theorem are not fully explored in relation to the inverse function theorem.

simpleeyelid
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Hello!

I am having problems with the inverse function theorem.

In some books it says to be locally inversible: first C1, 2nd Jacobian determinant different from 0

And I saw some books say to be locally inversible, it suffices to change the 2NDto "F'(a) is bijective"..

How could these two be equivalent.

Thank you for your kindness in advance,

Sincerely
 
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F'(a) is a linear map and its being bijective is (by linear algebra) equivalent to det F'(a) not being equal to zero.
 
Pere Callahan said:
F'(a) is a linear map and its being bijective is (by linear algebra) equivalent to det F'(a) not being equal to zero.

Thanks and, could you tell me the name of this proposition?
 
simpleeyelid said:
Thanks and, could you tell me the name of this proposition?

I'm not sure if this proposition has a special name. You may look up in wikipedia the equivalent conditions for a matrix to be invertible (which means that the associated linear map is an vector space isomorphism, i.e. bijective.)

http://en.wikipedia.org/wiki/Matrix_inversion"
 
Last edited by a moderator:
Pere Callahan said:
I'm not sure if this proposition has a special name. You may look up in wikipedia the equivalent conditions for a matrix to be invertible (which means that the associated linear map is an vector space isomorphism, i.e. bijective.)

http://en.wikipedia.org/wiki/Matrix_inversion"

MERCI beaucoup~~ I will check it..
 
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simpleeyelid said:
Thanks and, could you tell me the name of this proposition?
The general theorem is the "implicit function theorem" which basically says if the Jacobian of f(x,y,z) is not 0 at (x0, y0, z0) then we can solve for anyone of the variables as a function of the other two in some neighborhood of (x0, y0, z0).
 

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