Exercise 27, § 11

Given two topological spaces $X, Y$, one says that a mapping $f : X \to Y$ is a local homeomorphism if, for every $x \in X$, there exists a neighbourhood $U$ of $x$ such that $f(U)$ is a neighbourhood of $f(x)$ in $Y$ and the mapping $U \to f(U)$ coinciding with $f$ in $U$ is a homeomorphism. Every local homeomorphism is an open continuous mapping. a) Suppose that $X$ is separated, $Y$ locally connected. Let $f : X \to Y$ be a local homeomorphism for which there exists an integer $n > 0$ such that for every $y \in Y, f^{-1}(y)$ has exactly $n$ points. Now prove that for every $y \in Y$, there exists an open neighbourhood $V$ of $y$ such that $f^{-1}(V)$ has $n$ connected components $U_i$ ($1 \leq i \leq n$) and that the mapping $U_i \to V$ coinciding with $f$ in $U_i$ is a homeomorphism; $f$ is then a proper mapping. b) Suppose that $X$ is separated, $Y$ connected and locally connected. Let $f : X \to Y$ be a proper local homeomorphism; show that $f$ satisfies the assumption of a). (If, for every integer $n > 0$, $Y_n$ is the set of the $y \in Y$ such that $f(y)$ has at least $n$ points, show that $Y_n$ is both open and closed in $Y$). c) Give an example of a surjective local homeomorphism $f : X \to Y$, where $Y$ is compact, connected and locally connected, $X$ locally compact, connected and locally connected, $f^{-1}(y)$ has at most 2 points for every $y \in Y$, but $f$ is not proper (cf. I, p. 96, exerc. 4).