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Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is an analytic map. Can we say that there is a conull set $U \subset \mathbb{R}^n$ (i.e. $\mathbb{R}^n \setminus U$ has measure zero) where $|f^{-1}f(p)|=|f^{-1}f(q)|$ for all $p,q \in U$?

Any prod in the correct direction would be much appreciated as I am completely lost on where to even begin with a problem like this. Because of this, if any results are known for even the polynomial case I would be interested to know them.

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    $\begingroup$ Is $|f^{-1}f(p)|$ intended to mean the measure of the set $\{x: f(x) = f(p)\}$? $\endgroup$
    – Willie Wong
    Commented Nov 14, 2019 at 17:08
  • $\begingroup$ @WillieWong I am interested in the case where $|f^{-1}f(p)|$ is the cardinality of the set, not the measure. $\endgroup$
    – S. Dewar
    Commented Nov 14, 2019 at 19:56
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    $\begingroup$ Do you assume $m \geq n$? With $m < n$, the level sets are generically dimension $(n-m) > 0$ and and is uncountably infinite. $\endgroup$
    – Willie Wong
    Commented Nov 14, 2019 at 20:47
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    $\begingroup$ When $m \geq n$, the answer is no, by the way, since you can have simply functions like the polynomial $p: \mathbb{R}\to\mathbb{R}$ given by $p(x) = x^3 - x$. $\mathbb{R}$ splits into $K_1 \cup K_2 \cup K_3$, where $K_1 = (-\infty,-c)\cup (c,\infty)$, $K_2 = \{ \pm c\}$ and $K_3 = (-c,c)$ where $c = 1/\sqrt{3}$ (In this generality you also see that analyticity is not really important, since you are basically trying to describe index/Morse theory. ) $\endgroup$
    – Willie Wong
    Commented Nov 14, 2019 at 20:54
  • $\begingroup$ @WillieWong Thanks for the counter-example! Do you know any conditions that could be assumed so that my statement is true? Non-singular seems reasonable $\endgroup$
    – S. Dewar
    Commented Nov 15, 2019 at 13:35

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