We've just covered Sard's theorem and have just started to look at transversality in my differential geometry class and I'm trying to understand a proof of Sard's theorem (based on Milnor's proof): If $F:M^m \to N^n$ is a smooth map between smooth manifolds of dimension $m$ and $n$ respectively, then $F(C)$ has measure zero in $N$ (where $C = \{x \in M : {rank}\; dF_x < n\}$). The strategy behind the proof seems clear: write $C = (C\smallsetminus C_1)\cup (C_1 \smallsetminus C_2) \cup \cdots \cup (C_{k-1} \smallsetminus C_k) \cup C_k$ (where $C_i$ denotes the set of points in $M$ for which all partial derivatives of order $\le i$ vanish at $x$) and then show that the image under $F$ of each of the sets in the union on the r.h.s. has measure zero in $N$. But I'm a little confused on a few details and in particular I was wondering why each of the sets in the descending sequence $C \supset C_1 \supset C_2 \supset C_3 \supset \cdots$ is necessarily closed.

  • 2
    $\begingroup$ The $C_i$ are preimages of the closed set $\{0\}$ by continuous functions. $\endgroup$ – Pp.. Feb 3 '15 at 5:12
  • $\begingroup$ Thank you very much; I should probably always keep in mind this advantage of working in a smooth category :-). But, why is the set of critical points (C) itself closed in M? If it is the preimage of a closed set of N under a continuous map, then which set? $\endgroup$ – user212554 Feb 3 '15 at 13:05

Your Answer

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

Browse other questions tagged or ask your own question.