# About the Jacobian determinant and the bijection

1. Nov 21, 2008

### simpleeyelid

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.

Sincerely

Last edited: Nov 21, 2008
2. Nov 21, 2008

### Pere Callahan

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

3. Nov 21, 2008

### simpleeyelid

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

4. Nov 22, 2008

### Pere Callahan

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

5. Nov 22, 2008

### simpleeyelid

MERCI beaucoup~~ I will check it..

6. Nov 22, 2008

### HallsofIvy

Staff Emeritus
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 any one of the variables as a function of the other two in some neighborhood of (x0, y0, z0).