Proving that the number of elements in inversed sets are equal

Question

Define a cover mapping $f:Y\to X$ so that for all $x\in X$ the set $f^{-1}(x)$ is finite. Define a function $g:X\to \mathbb{Z}$ with $g(x) = \# (f^{-1}(x))$, as in: the number of elements in the set $f^{-1}(x)$.

I'm asked to proof that there is an open neighboorhood $U$ of $x$ in $X$ such that $g(x')=g(x)$ holds for all $x'\in U$.

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What I've got so far:

Let's take a random open neighboorhood $U$ of $x$ in $X$. I know that since $f$ is a cover mapping, $f^{-1}(U)$ is the distinct union of a family of open subsets of $Y$, and $f^{-1}(x) \subseteq f^{-1}(U)$ for all $x\in U$. I've got no clue how to go on from here. Can anyone provide some help?

Answer 1

Given $x\in X$, we find open $U\ni x$ such that $V:=f^{-1}(U)$ is of the form $V=\bigsqcup_{i\in I}U_i$ where $U_i\subseteq Y$ is open and $f|_{U_i}\colon U_i\to U$ is a homoeomorphism. Then $f^{-1}(x)$ intersects each $U_i$ exactly once, which means that $I$ has precisely $g(x)$ elements. Then for any $x'\in U$, we have that $f^{-1}(x')$ intersects each $U_i$ exactly once, which means that $g(x')=|I|=g(x)$.

Answer 2

EDIT: If you did not intend to state the definition of cover mapping, but just wanted to specify a cover mapping that additionally has finite fibers, then you may disregard this answer.

As you stated the definition of a "cover mapping", your statement is wrong. Let X be any space, and let $g: X \to \mathbb{N}$ be any map. Define Y to be disjoint union of $g(x), x \in X$, equipped with the discrete topology. Then the projection map $Y \to X$ is continous, because any map from a space which has the discrete topology is continous. But since $g$ was chosen arbitrarily, it will not in general have the property you hope for.

Try to reconsider what you mean by "cover mapping". For example, you could impose $f: Y \to X$ to be a local homeomorphism, or ask that for any $x \in X$ there is a neighbourhood $U_x$ such that $f^{-1}(U_i)$ is isomorphic to a disjoint union of copies of $U_i$. Then you could apply what Hagen said in his answer.