Is a bijective smooth function a diffeomorphism almost everywhere?

Solution 1:

I think the answer is no. Suppose $E\subset \mathbb R$ is closed, has positive measure, and has no interior (for example, $E$ could be the complement of an open set of small measure containing the rationals).

As is well known, there exists a $C^\infty$ function $f: \mathbb R\to [0,\infty)$ such that $f=0$ on $E$ and $f>0$ on $\mathbb R \setminus E.$ Define

$$F(x) = \int_0^x f(t)\, dt.$$

If $x<y,$ then $F(y) - F(x) = \int_x^y f.$ Because $f \ge 0$ and $[x,y]$ contains an interval in the complement of $E,$ this integral is $>0,$ hence $F(y) > F(x).$ Thus $F,$ which is $C^\infty,$ is strictly increasing, hence is a bijection onto $F(\mathbb R),$ a nice open interval. But $F'(x) = f(x)$ everywhere. Since $f= 0$ on $E,$ $F$ fails to be a local diffeomorphism at each point of $E,$ a set of positive measure.