Let $X$,$Y$ be compact, connected, smooth manifolds of the same dimension. Suppose you have a surjective smooth map $f : X \rightarrow Y$, such that $|f^{-1}(p) | \leq k$ for all $p \in Y$.

Let $U \subset X$ be an open dense subset.

Question: Is it possible to find $p \in Y$ such that $f^{-1}(p) \subset U$?


2 Answers 2


Yes. I think you do not even have to assume that the preimages are finite. $f$ can be any $C^1$ mapping from $X$ onto $Y$. Let $y\in Y$ be a regular value of the mapping $f$. Then for some $\epsilon>0$ $$ f^{-1}(B(y,\epsilon))=\bigcup_{i=1}^m U_i, $$ and $f|_{U_i}$ is a diffeomorphism of $U_i$ onto $B(y,\epsilon)$. $f(U\cap U_i)$ is open and sense in $B(y,\epsilon)$. If $$ p\in\bigcap_{i=1}^mf(U_i\cap U)\subset B(y,\epsilon), $$ then $f^{-1}(p)\subset U$.

  • 1
    $\begingroup$ It is worth mentioning that without the assumption of $X$ being compact, the assumption of finiteness of the preimages is needed to get the same $\varepsilon$ for all $U_i$. $\endgroup$
    – erz
    Mar 14, 2018 at 22:05

The answer is positive if we can assume that $U$ is Borel measurable (not necessarily open) and its complement is negligible. I will assume this in the sequel.

Fix Riemann metrics $g$ on $X$ and $h$ on $Y$. Denote by $k(p)$ the cardinality of $f^{-1}(y)$ and by $k_U(y)$ the cardinality of $f^{-1}(y)\cap U$. Denote by $J_f$ the Jacobian of $f$. More precisely for $x\in X$ we have

$$J_f(x)=\sqrt{\det D_xf\cdot(D_xf)^*}, $$

where $D_xf: T_x X\to T_{f(x)}Y$ is the differential of $f$ and $(D_x f)^*$ denotes its adjoint with respect to the inner products $g_x$ and $h_{f(x)}$.

The coarea formula shows that

$$ \int_Y k(y) dV_h(y)=\int_X J_f(x) dV_g(x)=\int_U J_f(x) dV_g(x). $$

The last equality follows from the fact that $X\setminus U$ has measure zero.

On the other hand the coarea formula (see Thm. 2.6 of this paper)shows that

$$\int_U J_f(x) dVg(x)=\int_Y|f^{-1}y)\cap U| dV_h(y)=\int_Y k_U(y) dV_h(y).$$


$$ \int_Y k(y) dV_h(y)= \int_Y k_U(y) dV_h(y) $$ or, equivalently

$$ \int_Y \Big(\; k(y)-k_U(y)\;\Big) dV_h(y)=0. $$

Since the measurable function $k(y)-k_U(y)$ is nonnegative we deduce from the above that $k(y)=k_U(y)$ for almost all $y$.


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