# How does this show one-to-one?

I'm reading a theorem in Munkres' "Analysis on Manifolds" that says

Let $A$ be an open subset of $\mathbb{R}^n$, let $f:A\to \mathbb{R}^n$ be of class $C^1$. If $Df(a)$ is non singular, then there exists $\alpha>0$ such that the inequality $$|f(x_0)-f(x_1)|\geq \alpha |x_0- x_1|$$

holds for all $x_0,x_1$ in an open cube $C(a;\epsilon)$ centered at $a$. It follows that $f$ is one-to-one on this open cube.

WHY does it follow from the conclusion that $f$ is one-to-one? Very basic question, but I don't see it. Thanks!

• If $x_0 \ne x_1$, what do you know about $\alpha|x_0 - x_1|$? – Anthony Carapetis Jun 12 '17 at 12:11

Let's assume that this function is not one-to-one. Then there exists at least one pair of points, $y_0, y_1$ s.t. $f(y_0)=f(y_1)$. Namely, this implies $f(y _0)-f(y_1)=0$, and $y_0-y_1 \ne 0$. Plugging this into the relation $$|f(y_0)-f(y_1)| = 0 \ge \alpha |y_0-y_1| > 0$$ Which is a contradiction.
If $f(x_0)=f(x_1)$, then $0\leqslant\alpha|x_0-x_1|\leqslant\bigl|f(x_0)-f(x_1)\bigr|=0$, and so $|x_0-x_1|=0$. Of course, this means that $x_0=x_1$.