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.


  • $\begingroup$ "except for a set of measure 0" in M or in N? ​ ​ $\endgroup$ – user57159 Jul 6 '16 at 19:20
  • $\begingroup$ 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. $\endgroup$ – Martin Arjovsky Jul 6 '16 at 20:19
  • $\begingroup$ 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.) $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.