# The Unapologetic Mathematician

## Another Lemma on Nonzero Jacobians

Sorry for the late post. I didn’t get a chance to get it up this morning before my flight.

Brace yourself. Just like last time we’ve got a messy technical lemma about what happens when the Jacobian determinant of a function is nonzero.

This time we’ll assume that $f:X\rightarrow\mathbb{R}^n$ is not only continuous, but continuously differentiable on a region $X\subseteq\mathbb{R}^n$. We also assume that the Jacobian $J_f(a)\neq0$ at some point $a\in X$. Then I say that there is some neighborhood $N$ of $a$ so that $f$ is injective on $N$.

First, we take $n$ points $\{z_i\}_{i=1}^n$ in $X$ and make a function of them

$\displaystyle h(z_1,\dots,z_n)=\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{x=z_i}\right)$

That is, we take the $j$th partial derivative of the $i$th component function and evaluate it at the $i$th sample point to make a matrix $\left(a_{ij}\right)$, and then we take the determinant of this matrix. As a particular value, we have

$\displaystyle h(a,\dots,a)=J_f(a)\neq0$

Since each partial derivative is continuous, and the determinant is a polynomial in its entries, this function is continuous where it’s defined. And so there’s some ball $N$ of $a$ so that if all the $z_i$ are in $N$ we have $h(z_1,\dots,z_n)\neq0$. We want to show that $f$ is injective on $N$.

So, let’s take two points $x$ and $y$ in $N$ so that $f(x)=f(y)$. Since the ball is convex, the line segment $[x,y]$ is completely contained within $N\subseteq X$, and so we can bring the mean value theorem to bear. For each component function we can write

$\displaystyle0=f^i(y)-f^i(x)=df^i(\xi_i)(y-x)=\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}(y^j-x^j)$

for some $\xi_i$ in $[x,y]\subseteq N$ (no summation here on $i$). But like last time we now have a linear system of equations described by an invertible matrix. Here the matrix has determinant

$\displaystyle\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{\xi_i}\right)=h(\xi_1,\dots,\xi_n)\neq0$

which is nonzero because all the $\xi_i$ are inside the ball $N$. Thus the only possible solution to the system of equations is $x^i=y^i$. And so if $f(x)=f(y)$ for points within the ball $N$, we must have $x=y$, and thus $f$ is injective.

November 17, 2009 - Posted by | Analysis, Calculus