Results 1 to 2 of 2

Math Help - Injective

  1. #1
    Aug 2009


    Let \subseteq \mathbb{R}^2" alt="f \subseteq \mathbb{R}^2" /> \to \mathbb{R}^2. Show that if the determinant of the Jacobian Matrix of f is not equal to zero at (x_o,y_o)\in D, then f is injective in some neighborhood of (x_o,y_o).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Apr 2005
    First, the Jacobian is continuous in (x, y) (you may need to show that) so if it is not 0 at (x_0, y_0) it is not 0 in some neighborhood of that point. Now use the "mean value theorem". If f(x_1, y_1)= f(x_2, y_2) then f(x_1, y_1)- f(x_2, y_2)= J(x, y)\begin{bmatrix}x \\ y\end{bmatrix}= 0 for some point (x, y) on the line between (x_1, y_1) and (x_2, y_2)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. f: [0,1]-->]0,1[, f injective
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 1st 2010, 09:08 AM
  2. If gof is injective, then f is injective
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 17th 2009, 11:10 PM
  3. B & AB injective but A not injective
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 24th 2009, 10:33 PM
  4. g(f(x) is injective, is f injective?
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: November 9th 2008, 07:23 PM
  5. Injective
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 23rd 2006, 11:09 AM

Search Tags

/mathhelpforum @mathhelpforum