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The fixed-locus Borel base is paracompact and of CW type

Statement

Assume AC. Let r≥0, let T=(R/Z)r use the product Milnor model BT=(CP∞)r with its ordinary product topology, and let F be a compact finite-dimensional Hausdorff second-countable smooth manifold without boundary. Then BT×F is paracompact Hausdorff and has the homotopy type of a CW complex. In fact, there is a finite CW complex K such that BT×F≃BT×K.

Statement references

Facts & Assumptions

Given: AC, r≥0, the product-torus classifying-space model, and compact smooth F.

[A1]

Full AC includes the choice hypotheses of the compact-manifold and product classifying-space constructions. The Axiom of Choice

[F1]

For r≥1, BT=(CP∞)r has a countable CW structure whose CW topology is the ordinary product topology; the complete product-topology argument is given in the supplier's proof. The universal complex flag bundle is BT-n

[F2]

A compact smooth Riemannian manifold admits a finite strongly geodesically convex cover whose finite nerve is homotopy equivalent to it. A compact smooth manifold has a finite good-cover nerve model

[F3]

Every CW complex and every compact Hausdorff space is paracompact Hausdorff. Hatcher's proof for CW complexes constructs subordinate partitions over the skeleta; the compact Hausdorff case is his Proposition 1.18. [Hatcher, Appendix, Propositions 1.18 and 1.20, printed pp. 35–37.]

Proof

Given: The data in the statement.

1.1A1F1F2F3construct

If r=0, BT is a point and the CW and paracompact assertions reduce to the compact manifold F. For r≥1, [F1] gives BT its ordinary product CW topology, and [F3] makes it paracompact Hausdorff. By [F2], choose a finite CW complex K and a homotopy equivalence h:F→K. Since K is finite, its CW product with the CW complex BT has the ordinary product topology; hence BT×K is a CW complex. The maps idBT×h and the product of a homotopy inverse with idBT are homotopy inverses, since their coordinate homotopies are continuous in the product topology. Thus BT×F has CW type. If F=∅, take K=∅.

2.1F1F3givenconstruct∎

If F=∅, the product is empty and paracompact Hausdorff, so assume F≠∅. To prove paracompactness of BT×F, let U be an open cover. For each x∈BT and y∈F, choose a product neighborhood Ox,y×Vx,y of (x,y) contained in some member Ux,y∈U. Compactness of F selects finitely many y1,…,ym whose Vx,yj cover F. Then Ox=⋂j=1mOx,yj is an open neighborhood of x, and Ox×F is covered by Ux,y1,…,Ux,ym. Paracompactness of BT gives a locally finite open refinement {Wα} of {Ox}. For each α, choose xα with Wα⊆Oxα; the finitely many sets (Wα×F)∩Uxα,yj cover Wα×F, are open, and refine U. They are locally finite because a neighborhood N×F meets sets indexed by only those α for which Wα meets N, a finite set. Hence BT×F is paracompact. It is Hausdorff as a product of Hausdorff spaces. The point and rank-zero cases satisfy the same conclusions.

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