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Leray–Hirsch fails without a global restricting fiber basis

Statement refuted

Assume AC. It is false that free constant-rank fiber cohomology alone, without global classes restricting to a fiber basis, gives the Leray–Hirsch module isomorphism. For the Klein-bottle bundle

$S^{1} ⟶ K = T_{r} ⟶ S^{1}; r (z) = \overline{z},$

reflection monodromy prevents a global integral fiber generator and $H^{1} (K; Z) ≅ Z$ , not the rank-two group predicted by treating the fiber basis as constant.

Facts & Assumptions

Given: AC, integral coefficients, the counterclockwise orientation of $S^{1}$ , and the displayed reflection mapping torus.

[F1]

A global fiber basis trivializes Serre monodromy says that global classes restricting to a fiber basis force cohomological fiber transport to fix that named basis.

[F2]

Degree of identity constant reflection and antipodal sphere maps says a circle reflection has degree $- 1$ .

[F3]

Wang sequence for a fibration over the circle gives the integral homology sequence with maps $1 - T_{q}$ .

[F4]

Homology of spheres gives $H_{0} (S^{1}; Z) = H_{1} (S^{1}; Z) = Z$ and zero homology in higher degrees.

[F5]

Topological universal coefficient short exact sequence for cohomology gives the integral cohomology evaluation sequence under AC.

[A1]

The Axiom of Choice is used exactly in [F1] and [F5].

Counterexample

technique · calculate the reflection monodromy, then compare the actual and falsely untwisted degree-one groups

1.1F2F4

Write $K = (S^{1} \times [0, 1]) / (z, 1) \sim (r (z), 0)$ . Product charts away from the seam and charts changing fiber coordinate by $r = r^{- 1}$ across the seam make $K \to S^{1}$ a fiber bundle. Positive-loop transport is $r$ , so [F2] and [F4] give $T_{1} = - 1$ on $H_{1} (S^{1}; Z) = Z$ and $T_{0} = 1$ on $H_{0} (S^{1}; Z) = Z$ .

2.1F1F5step 1.1

Cohomological transport on $H^{1} (S^{1}; Z)$ is likewise multiplication by $- 1$ , since evaluation on the homology generator changes by the degree in step 1.1. It fixes no generator. The contrapositive of [F1] therefore says that no global class on $K$ can restrict to an integral basis of $H^{*} (S^{1}; Z)$ .

2.2F3F4step 1.1

The degree-one part of [F3], using step 1.1, gives $0 \to Z / 2 \to H_{1} (K; Z) \to Z \to 0$ . The fixed point $1 \in S^{1}$ defines the section $[t] \mapsto [(1, t)]$ , so the projection onto the last $Z$ splits and $H_{1} (K; Z) ≅ Z \oplus Z / 2$ . The same explicit mapping-torus model is path connected, hence $H_{0} (K; Z) = Z$ .

3.1F5A1step 2.2

Apply [F5] in degree one. Since $H_{0} (K) = Z$ is free, its Ext term is zero, and every homomorphism $Z / 2 \to Z$ is zero. Consequently $H^{1} (K; Z) ≅ Hom (Z \oplus Z / 2, Z) ≅ Z$ .

4.1F4F5step 2.1step 3.1

By [F4] and [F5], both base and fiber have one copy of $Z$ in cohomological degrees zero and one. Falsely declaring the fiber basis $(1, u)$ constant would make the degree-one Leray–Hirsch source $H^{1} (S^{1}) \otimes ⟨ 1 ⟩ \oplus H^{0} (S^{1}) \otimes ⟨ u ⟩ ≅ Z^{2}$ , whereas step 3.1 gives only $Z$ . The failed conclusion and its missing global-basis hypothesis are therefore witnessed explicitly.

5.1F1F2F3F4F5A1step 1.1step 2.1step 2.2step 3.1step 4.1∎

The base, fiber and total space are nonempty, and the coefficient ring is fixed as nonzero $Z$ . Steps 1.1–4.1 include the one base loop, its two seam endpoints, the degree-zero unit, the zero kernel of multiplication by two, identity action on $H_{0}$ , reflection action on $H_{1}$ , and the degenerate false identity-monodromy comparison. AC is used only through [A1] in [F1] and [F5]; the mapping-torus and Wang calculations are choice-free. No converse claim is made.

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Sources

Miller, MIT 18.906 notes, Lecture 33 Leray–Hirsch hypotheses (standard reference, not scraped)