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Basic stability of proper continuous maps

Statement

Let X, Y and Z be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with properness as in Proper continuous maps of topological spaces.

  1. If i:Z↪X is the inclusion of a closed subset, then i is proper.
  2. If f:X→Y and g:Y→Z are proper continuous maps, then the composite gf:X→Z is proper.
  3. If f:X→Y is proper, Y is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and u:Y′→Y is continuous, then the pullback f′:X×YY′⟶Y′,X×YY′={(x,y′)∈X×Y′:f(x)=u(y′)}, the subset carrying the subspace topology of the product (The product set ∏i∈IXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), is proper.
  4. If f:X→Y is proper, then every fibre f−1(y) is a compact subset of X.
  5. If f:X→Y is proper and S⊆X is closed, then the restriction f∣S:S→Y is proper.
  6. If f:X→Y is proper and Y is locally compact Hausdorff (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space), then f is a closed map: the image of every closed subset of X is closed in Y.

These are the stability properties consumed by the closed-support functor f! on this page; no choice principle is used in any of them.

Facts & Assumptions

Given: Topological spaces X, Y, Z and Y′, continuous maps as in the Statement, and the intrinsic convention for compactness of subsets.

[F1]

A continuous f:X→Y is proper exactly when f−1(K) is a compact subset of X for every compact subset K⊆Y (Proper continuous maps of topological spaces).

[F3]

A closed subset of a compact space is a compact subset of it, and finitely many compact subsets have compact union (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

[F4]

A finite product of compact spaces is compact; in particular the binary product of two compact spaces is compact, and a one-point space is compact (A product of finitely many compact spaces is compact in the product topology).

Proof

technique · direct
1.1givenF1F3F6

Closed inclusions are proper. Let i:Z↪X be the inclusion of a closed subset and let K⊆X be compact. Its preimage K∩Z is closed in the compact space K, hence compact by [F3]. The topology on K∩Z induced from Z is the same as that induced from K or X, by transitivity of subspace topology [F6]. Thus i−1(K) is compact in Z, and i is proper by [F1].

1.2givenF1algebra

Composites of proper maps are proper. For K⊆Z compact, (gf)−1(K)=f−1(g−1(K)); the set g−1(K) is compact by [F1] applied to g, and then its f-preimage is compact by [F1] applied to f. Hence gf is proper.

1.3givenF1F4

Fibres of a proper map are compact. A one-point space is compact by [F4], so for each y∈Y the preimage f−1({y}) is compact by [F1].

1.4givenF1F3F6

Restrictions to closed subspaces are proper. Let S⊆X be closed, and give S its subspace topology. For a compact K⊆Y one has (f∣S)−1(K)=f−1(K)∩S. The set f−1(K) is compact by [F1], and f−1(K)∩S is closed in the compact space f−1(K) because S is closed in X; hence f−1(K)∩S is a compact subset of f−1(K) by [F3]. Its topology as a subset of S agrees with its topology as a subset of X by [F6], so (f∣S)−1(K) is a compact subset of S and f∣S is proper by [F1].

1.5givenF1F2F3F4F5F6F7

Properness is preserved by base change along a map into a Hausdorff space. Let K⊆Y′ be compact and put K0=u[K], a compact subset of Y by [F2]. Then f−1(K0) is compact by [F1], so the product f−1(K0)×K is a compact subspace of X×Y′ by [F4]. The pairing (f,u):X×Y′→Y×Y is continuous by [F7], and ΔY is closed in Y×Y by [F5]; hence X×YY′=(f,u)−1(ΔY) is closed in X×Y′, and its intersection with the subspace f−1(K0)×K is closed in that compact space. Moreover this intersection is exactly (f′)−1(K): a pair (x,y′) lies in X×YY′ with y′∈K precisely when f(x)=u(y′) and y′∈K, and then u(y′)∈K0 forces x∈f−1(K0), so (f′)−1(K)=X×YY′∩(f−1(K0)×K). By [F3] this is a compact subset of the product X×Y′, and by [F6] its topology as a subset of X×YY′ agrees with that as a subset of X×Y′; hence f′ is proper by [F1].

1.6givenF1F2F3F8

Proper maps into locally compact Hausdorff spaces are closed. Let f:X→Y be proper with Y locally compact Hausdorff, let F⊆X be closed and let y∈Y∖f[F]. By [F8] there is an open neighbourhood V of y whose closure V‾ is compact. Then f−1(V‾) is compact by [F1] and F∩f−1(V‾) is closed in it, hence compact by [F3]; its image under the continuous map f is f[F]∩V‾, which is compact by [F2] and closed in Y by [F8]. Therefore V∖(f[F]∩V‾) is open in Y, contains y because y∈V and y∉f[F], and is disjoint from f[F]: a point of it lies in V, hence cannot lie in f[F]∖V‾, and it is disjoint from f[F]∩V‾ by construction. Hence f[F] is closed.

2.1givenstep 1.1step 1.2step 1.3step 1.4step 1.5step 1.6∎

The six stability properties are established: claim 1 by step 1.1, claim 2 by step 1.2, claim 4 by step 1.3, claim 5 by step 1.4, claim 3 by step 1.5 and claim 6 by step 1.6. Every argument uses only the preimage form of properness [F1], the preservation of compactness under continuous maps [F2], closed-in-compact compactness [F3], finite products of compact spaces [F4], the closed-diagonal criterion [F5], the intrinsic subspace conventions [F6], the initial product topology [F7] and the locally compact Hausdorff basis of compact neighbourhoods [F8]; none of these selects points or open sets from an infinite family, so no choice principle is used.

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