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Leray–Hirsch for a trivial product bundle

Example

Assume AC. Let $B$ be a path-connected CW complex, let $R$ be a commutative PID, and suppose every $H_{q} (F; R)$ is finite free and finitely many homogeneous classes $b_{1}, \dots, b_{m}$ form an $R$ -basis of $H^{*} (F; R)$ . For $p : B \times F \to B$ , the classes $1 \times b_{i}$ give Leray–Hirsch and recover the cohomological Künneth module isomorphism.

Facts & Assumptions

Given: The spaces, PID and finite homogeneous basis above.

[F1]

Cohomological Kunneth isomorphism under finite free hypotheses gives $H^{*} (B \times F; R) ≅ H^{*} (B; R) \otimes_{R} H^{*} (F; R)$ by external product, under AC and the stated finite-free fiber homology hypothesis.

[F2]

Leray–Hirsch module isomorphism gives the module isomorphism from a supplied global restricting fiber basis.

[A1]

The Axiom of Choice is used exactly through [F1]–[F2].

Verification

technique · calculate restrictions and the displayed formula

1.1

Define $e_{i} = 1 \times b_{i} = {pr}_{B}^{*} 1 ⌣ {pr}_{F}^{*} b_{i}$ . On the fiber ${b} \times F$ , the first factor restricts to $1$ and the second to $b_{i}$ , so $e_{i} ∣_{F_{b}} = b_{i}$ . The supplied list is therefore a basis on every fiber.

2.1

Apply [F2]. Its map sends $(a_{i})$ to $\sum_{i} p^{*} a_{i} ⌣ e_{i} = \sum_{i} a_{i} \times b_{i}$ . This is exactly the cross-product map in [F1], and the basis identifies its source with $H^{*} (B; R) \otimes_{R} H^{*} (F; R)$ . Thus both isomorphisms agree, not merely their abstract modules.

F1F2step 1.1

3.1

If $m = 0$ , the fiber cohomology and both sides are zero; $m = 1$ gives one shifted copy. A point fiber has basis $1$ , and a point base recovers $H^{*} (F)$ . Empty $F$ , degree zero, zero classes and all finite direct-sum endpoints are included in the formulas. The zero ring is outside the stated PID convention. No basis is chosen: it is supplied. AC is used exactly through [A1] in Künneth and cohomological Serre/Leray–Hirsch.

F1F2A1step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Miller, MIT 18.906 notes, Lecture 33 (standard reference, not scraped)