Let $Y$ be a locally compact Hausdorff space, and $X$ a compact Hausdorff space (we can assume they are metric spaces if needed). Suppose that $p : Y \to X$ is a local homeomorphism for which there exists $N \in \mathbb{N}$ such that $\# p^{-1}(x) \le N$ for all $x \in X$.

Let $C_0(Y)$ denote the continuous, complex-valued functions on $Y$ which "vanish at infinity". That is, $f \in C_0(Y)$ if $f$ is continuous and for all $\varepsilon > 0$ there exists a compact set $K \subseteq Y$ such that $|f(y)| < \varepsilon$ for all $y \in Y \setminus K$.

Now fix $f \in C_0(Y)$. Is the function $x \mapsto \sum_{y \in p^{-1}(x)} f(y)$ continuous on $X$?


Let $Y$ be the subset of $\mathbb{R}^2$ given by $Y = \{(x,1) \mid x \in [0,1]\} \cup \{ (x,2) \mid x \in [0,1/2)\}.$ Let $X = [0,1]$ and define a local homeomorphism $p : Y \to X$ by $p(x,y) = x$. It appears as though $x \mapsto \sum_{y \in p^{-1}(x)} f(y)$ is continuous for $f \in C_0(Y)$, however I can't for the life of me work out how to write down a proof. It seems to rely heavily on the fact that $f$ vanishes at $(1/2,2)$.


1 Answer 1


We prove the required claim under no assumptions for $X$ (other than Hausdorffness) and $Y$. For each $x\in X$ put $$P(x)=\sum_{y \in p^{-1}(x)} f(y)$$ (if $x\not\in\operatorname{Im} f$ then we shall treat such sums as zeros). Assume that the map $P$ is discontinuous. There exists a point $x\in X$ and a number $\varepsilon>0$ such that for each neighborhood $U$ of the point $x$ there exists a point $x_U\in U$ such that $|P(x_U)-P(x)|\ge 2N\varepsilon$. There exists a compact set $K\subset Y$ such that $|f(y)|<\varepsilon$ for all $y\in Y\setminus K$ (I assume that this should be in your definition instead of $y\not\in Y\setminus K$). Also, since the function $p$ is a local homeomorphism, each point $y\in p^{-1}(x)$ has an open neighborhood $V_y’$ such that the map $y’\mapsto p(y’)$ is a homeomorphism between $V_y’$ and an open subset $p(V_y’)$ of $X$. Next, since the function $f$ is continuous, for each point $y\in p^{-1}(x)$ there exists its open neighborhood $V_y\subset V_y’$ such that $|f(y’)-f(y)|<\varepsilon$ for each point $y’\in V_y$. Put $$K’=K\setminus\bigcup\{V_y: y\in p^{-1}(x)\}.$$ Then $K’$ is a compact and $p(K')\not\ni x$. Since the space $X$ is Hausdorff, the set $p(K’)$ is closed. Thus the set

$$U=\cases {\bigcap\{ p(V_y): y\in p^{-1}(x)\}\setminus p(K’), \mbox{ if } x\in\operatorname{Im} f,\\ X\setminus p(K’), \mbox{ if } x\not\in\operatorname{Im} f.}$$ is an open neighborhood of the point $x$. So there exists a point $x_U\in U$ such that $$|P(x_U)-P(x)|\ge 2N\varepsilon.$$ Let $y\in p^{-1}(x)$. Since $U\subset p(V_y)\subset p(V_y’)$, there exists a unique point $y_U\in V_y$ such that $p(y_U)=x_U$. Put $$A=p^{-1}(x_U)\setminus \{y_U:y\in p^{-1}(x)\}.$$ Remark that $A$ is disjoint from $K$.


$$|P(x_U)-P(x)|=$$ $$\left| \sum_{y \in p^{-1}(x_U)} f(y) - \sum_{y \in p^{-1}(x)} f(y)\right|=$$ $$\left|\sum_{y \in A} f(y)+ \sum_{y \in p^{-1}(x)} f(y_U) - \sum_{y \in p^{-1}(x)} f(y)\right|\le$$ $$\sum_{y \in A} |f(y)|+\sum_{y \in p^{-1}(x)} |f(y_U)- f(y)|\le $$ $$\sum_{y \in Y\setminus K} |f(y)|+\sum_{y \in p^{-1}(x)} |f(y_U)- f(y)|<$$ $$N\varepsilon +N\varepsilon,$$

a contradiction.


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