Prove that $|f|\leq 1$ whenever $|x|\leq 1$. 
Let $f :\mathbb{R}^2\rightarrow \mathbb{R}^2 $ be everywhere differentiable such that the Jacobian is not singular at any point in $\mathbb{R}^2$. Assume $|f|\leq 1$ whenever $|x|=1$. Prove that $|f|\leq 1$ whenever $|x|\leq 1$.

I think this is straight forward if we apply maximum modulus principle. But how may I prove it without using it? I tried to use Implicit function theorem by defining $g:\mathbb{R}^2\rightarrow \mathbb{R}$ by $g(x)=|f(x)|^2$, but no success.
At least a hint is appreciated.
 A: Consider $g(x) = |f(x)|^2$. We note by the chain rule that
$$
Dg(x) = 2[f(x)]^T Df(x)
$$
In order for $g$ to achieve a maximum at $x$ on the interior of the unit ball, it must have a critical point at that same $x$, i.e. a point such that $Dg(x) = 0$. Because the Jacobian $Df$ is non-singular at all points, this would imply that $f(x) = 0$.
So, $g(x) = 0$ at any potential maximum on the interior. So, $g$ has no local maximum on the interior of the unit ball. The conclusion follows.
A: Let $x_0$ be a point in the disk $D=\{|x|\le 1\}$ where $|f|$ has an absolute maximum. Suppose by contradiction that $|f(x_0)|>1$. Then $x_0$ is an internal point of $D$ (because on the boundary $|f|=1$. Hence, by assumption, $Df_{x_0}$ is invertible, and by the local invertibility Theorem the function $f$ is invertible in a neighbourhood $U$ of $x_0$ all contained in $D$. Hence $f(D) \supset f(U) \supset V$ where $V$ is a neighborhood of $f(x_0)$. But this is a contradiction, because in $V$ there are points $y$ with $|y|$ larger than $|f(x_0)|$.
A: The set $C=\{x | \|x\| \le 1 \}$ is compact, hence there is a maximiser $x_0 \in C$ of
of $|f|$.
If $|f(x_0)| \le 1$ then there is nothing to show, so suppose
$|f(x_0)| >1$, in which case we must have $x_0 \in \{x | \|x\| < 1 \}$
and hence $\phi(x)=|f(x)|$ has a local maximiser at $x_0$. Since
$\phi$ is differentiable at $x_0$, we have
$D \phi(x_0) = {f(x_0) \over |f(x_0)| } D f(x_0) = 0$ we see that
$Df(x_0) = 0$ which is a contradiction.
