Proper Maps and Universally Closed Maps Reinhard Schultz Continuous functions defined on compact spaces generally have special properties For example if X is a compact space and Y is a Hausdorff space then f is a closed mapping There is a useful class of continuous mappings called proper perfect or compact maps that satisfy many properties of continuous maps with compact domains In this note we shall study a few basic properties of these maps Further information about proper maps can be found in Bourbaki B2 Section I 10 Dugundji D Sections XI 5 6 pages 235 240 and Kasriel K Sections 95 and 105 pages 214 217 and 243 247 There is a slight difference between the notions of perfect and compact mappings in B2 D and K and the notion of proper map considered here Specifically the definitions in D and K include a requirement that the maps in question be surjective The definition in B2 is entirely different and corresponds to the notion of universally closed map discussed later in these notes the equivalence of this definition with ours for a reasonable class of spaces is proved both in B 2 and in these notes see the section Universally closed maps PROPER MAPS If f X Y is a continuous map of topological spaces then f is said to be proper if for each compact subset K Y the inverse image f 1 K is also compact EXAMPLE Suppose that A is a finite subset of R n and f Rn A Rm is a continuous function such that for all a A we have lim f x lim f x x a x Then f is proper Proof If K Rm is compact then K is contained in some large disk D The limit condition at implies that f 1 D is contained in some large disk D 0 Rn furthermore the limit conditions at all the points ai A imply that f 1 D is also contained in the complement of E D 0 Ni where Ni is a small open disk neighborhood with center a i But E is compact and f 1 K is a closed subset of Rn that is contained in E it follows that f 1 K is compact IMPORTANT SPECIAL CASE Suppose n m 2 and f z p z q z where p and q are complex polynomials with deg p deg q Then f satisfies the limit condition and therefore is proper The first result which is fairly standard shows that proper maps of certain not necessarily compact spaces have some important properties in common with continuous maps of compact Hausdorff spaces 1 2 THEOREM 1 Suppose that X and Y are Hausdorff spaces with Y either locally compact or metrizable and let f X Y be continuous Then f is proper if and only if f is closed and for each y Y the set f 1 y is compact Proof Suppose that f is closed and inverse images of one point sets are compact we claim that inverse images of arbitrary compact subsets are compact Let K be a compact subset of Y and let F F be a collection of closed subsets of f K with the finite intersection property Without loss of generality we may assume that F is closed under finite intersections for if F 0 is the family of all finite intersections of sets in F then F 0 is closed under finite intersections and the intersection of all the sets in F equals the intersection of all the sets in F 0 Since f is closed the subsets f F are closed in Y since F f 1 K is true by assumption it follows that f F K Combining these observations we see that f F is a compact subset of K But the family of closed sets f F also has the finite intersection property because 1 6 f F 1 f F k f F 1 f F k Therefore 6 f F by compactness of K Let y be a point in the intersection ily F is closed under finite intersections for all 1 n we have f 1 y F 1 Therefore the family f 1 y F has the finite intersection property is compact and therefore 6 f 1 y F F Therefore the compact Since the fam F n 6 But f 1 y set f 1 K is Suppose now that f is proper and Y is locally compact or metrizable If F X is a closed subset then it is immediate that f F is proper Therefore it suffices to prove that if f is proper then f X is closed in Y because f F f F F Assume first that Y is locally compact Let y be a point in the closure of f X and let K be a compact neighborhood of y Then f 1 K is a nonempty compact set and f f 1 K a closed mapping Therefore f f 1 K f X K is a closed set But by construction y is a limit point of this set and consequently y f X K f X Assume now that Y is a metric space Let F be closed in X and let x n be a sequence of points in F such that lim f xn y Let Cn be the compact set y f xn f xn 1 so that f 1 Cn is a nonempty nested sequence of compact sets Therefore by compactness we have that 6 n f 1 Cn F f 1 n Cn F f 1 y F Therefore y f F must hold and consequently f F must be closed in Y Remark The implication does not require an extra condition on Y The implication is valid more generally if Y is a k space see D Section XI 9 especially XI 9 3 on p 248 Here is the proof that f is closed under this hypothesis The set f X is closed if and only if f X K is compact for all compact subsets K Y But f X K f f 1 K and this set is compact because f proper f 1 K is compact and f 1 K compact f f 1 K compact THEOREM 2 Suppose that f X Y is a proper map and B Y Then f f 1 B is proper Conversely if B is either an open covering or a finite closed covering of Y and each of the maps f f 1 B is proper then f is also proper 3 Proof Let fB f f 1 B If K is a compact subset of B then f B 1 K f 1 K but the latter is compact since f is proper and therefore it follows that f B is proper Let f f f 1 B and let K Y be compact Suppose that B is an open covering of Y Then by compactness K is contained in a finite union B 1 B k Let V j K B j then the …