proper maps to locally compact spaces are closed



topology (point-set topology, point-free topology)

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(locally compact topological space)

A topological space XX is called locally compact if for every point xXx \in X and every open neighbourhood U x{x}U_x \supset \{x\} there exists a smaller open neighbourhood V xU xV_x \subset U_x whose topological closure is compact and still contained in UU:

{x}V xCl(V x)compactU x. \{x\} \subset V_x \subset \underset{\text{compact}}{Cl(V_x)} \subset U_x \,.

Statement and proof


(proper maps to locally compact spaces are closed)


  1. (X,τ X)(X,\tau_X) be a topological space,

  2. (Y,τ Y)(Y,\tau_Y) a locally compact topological space according to def. 1,

  3. f:XYf \colon X \to Y a continuous function.


If ff is a proper map, then it is a closed map.


Let CXC \subset X be a closed subset. We need to show that every yY\f(C)y \in Y \backslash f(C) has an open neighbourhood U y{y}U_y \supset \{y\} not intersecting f(C)f(C) (by this prop.).

By local compactness of (Y,τ Y)(Y,\tau_Y) (def. 1), yy has an open neighbourhood V yV_y whose topological closure Cl(V y)Cl(V_y) is compact. Hence since ff is proper, also f 1(Cl(V y))Xf^{-1}(Cl(V_y)) \subset X is compact. Then also the intersection Cf 1(Cl(V y))C \cap f^{-1}(Cl(V_y)) is compact, and hence so is

f(Cf 1(Cl(V y)))=f(C)(Cl(V))Y. f(C \cap f^{-1}(Cl(V_y))) = f(C) \cap (Cl(V)) \; \subset Y \,.

This is also a closed subset, since compact subspaces of Hausdorff spaces are closed. Therefore

U yV y\(f(C)(Cl(V y)))=V y\f(C) U_y \coloneqq V_y \backslash ( f(C) \cap (Cl(V_y)) ) = V_y \backslash f(C)

is an open neighbourhod of yy not intersecting f(C)f(C).

Revised on May 16, 2017 13:49:24 by Urs Schreiber (