Is a bijective smooth function a diffeomorphism almost everywhere?

Suppose I have $f: M \rightarrow N \in C^{\infty}$ a smooth bijection between $n$-dimensional smooth manifolds. Does it have to be a diffeomorphism except for a set of measure 0?

I think the proof might come from showing that $X = \{p: d_pf \text{ is not an isomorphism}\}$ has measure zero. Using the inverse function theorem you can show that the statement follows from this. By Sard's theorem, we know that $f(X)$ has measure zero, but I don't know how to go from there to $X$ having measure zero (since we don't know, for example, that $f^{-1}$ is locally Lipschitz).

You may assume (if you want) that $M$ and/or $N$ are connected and/or compact.

Thanks!

• "except for a set of measure 0" in M or in N? ​ ​ – user57159 Jul 6 '16 at 19:20
• In $M$. Meaning the set $Z = \{p: \text{there does not exist an open set$U$that contains$p$on which$f|_U$is a diffeomorphism to it's image}\}$ has measure 0. Note that since $f$ is smooth this implies the same for $f^{-1}$ because $f(Z)$ would have measure zero. – Martin Arjovsky Jul 6 '16 at 20:19
• The smooth restriction is significant, and probably makes this true. I can break it if I'm allowed to be non-smooth (but continuous!) at a set of measure zero, and a diffeomorphism elsewhere. Indeed apply uniformization on the interior of a Jordan curve of positive measure and then Caratheodary's theorem, and then take the inverse of this. (Do this to both sides of the circle.) – user98602 Jul 6 '16 at 21:36

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.

• Very nice, looks like my guess was totally off. – user98602 Jul 6 '16 at 23:09
• Thanks! I think your argument is correct. – Martin Arjovsky Jul 7 '16 at 0:15