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Consider $f: \mathbb{R}^2 \to \mathbb{R}^2$, where $f$ is given by

$$f =\begin{bmatrix} f_1(x,y)\\ f_2(x,y) \end{bmatrix}$$

Where we may assume $f$ is of class $C^1$ (continuously differentiable), if $J$ is the associated Jacobian of $f$ and if $\det J \neq 0$ for every $x,y \in\mathbb{R}^2$, why does that imply $f$ is one to one?

I understand this idea for single variable functions, but not so convinced in multivariables.

EDIT,

Thank you for the counterexample, but let me add one more restriction. Let $f$ be one to one defined on an open set $A$, so $f: A \to \mathbb{R}^2$

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2 Answers 2

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The best you can say is that there is a neighborhood on which your function is one-to-one; this is the Inverse Function Theorem, which applies when the Jacobian is non-degenerate.

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  • $\begingroup$ So just locally one-to-one? But the Inverse Function Theorem seem to only apply to $\mathbb{R}^n \to \mathbb{R}^n$ $\endgroup$
    – Jennifer
    Mar 5, 2013 at 5:18
  • $\begingroup$ Well, your case is $n=2$ @Jennifer $\endgroup$
    – Pedro
    Mar 5, 2013 at 5:24
  • $\begingroup$ Is there a version if $\mathbb{R}^m \to \mathbb{R}^n$ is what I to know. But it may be dumb because an inverse may not even exist if the dimensions are different $\endgroup$
    – Jennifer
    Mar 5, 2013 at 5:27
  • $\begingroup$ @Jennifer There's the implicit function theorem. Both the inverse and implicit function theorems are specific cases of a more general theorem. en.wikipedia.org/wiki/… $\endgroup$
    – dls
    Mar 5, 2013 at 5:33
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This is false. Consider the function $f(x,y)=[e^x\cos(y),e^x\sin(y)]$. Compute the determinant of the Jacobian to see that it never vanishes. However, the function is $2\pi$-periodic in the $y$ variable, that is, $f(x,y)=f(x,y+2\pi k)$ for any integer $k$.

Side note: if you know about complex variables, then this function may be written as $f(z)=e^z$.

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  • $\begingroup$ @dls, actually out of curiosity, what does the transformation of $f$ do? Is it all of $\mathbb{R}^2$? If I remove $x = 0$, then $f$ maps without the unit circle right (a trace of the unit circle is missing in other words) $\endgroup$
    – Jennifer
    Mar 5, 2013 at 5:17
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    $\begingroup$ Fix $x=3$. Then $f(3,y)=e^3[\cos(y),\sin(y)]$, so the image of this vertical line is a circle of radius $e^3$ centered at the origin. Say we fix $y=\pi/2$, a horizontal line. Then $f(x,\pi/2)=e^x[0,1]$, the image of the horizontal line $y=\pi/2$ is the line through the origin making angle $\pi/2$ with the positive $x$-axis. $\endgroup$
    – dls
    Mar 5, 2013 at 5:24
  • $\begingroup$ Sorry, that should be "ray" starting at origin... not line through the origin. In fact, the point $[0,0]$ is not in the image of $f$. So $f$ is not onto. (But this is the only point that's missed.) $\endgroup$
    – dls
    Mar 5, 2013 at 5:30
  • $\begingroup$ So suppose $0 < y < 2\pi$, would the image be the all of $\mathbb{R}^2$, but the entire x-axis since I get every possible choice of circles $\endgroup$
    – Jennifer
    Mar 5, 2013 at 5:31
  • $\begingroup$ Almost! (I led you astray with my first comment.) We can never get the "zero radius circle" since $e^x>0$ for every $x$. The image of such a "horizontal strip" equals $\mathbb{R}^2-\{[0,0]\}$. Also, you should really include either $y=0$ or $y=2\pi$ otherwise you're omitting the positive $x$-axis, as well. $\endgroup$
    – dls
    Mar 5, 2013 at 5:36

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