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A closed null fence word has an essential lower endpoint

Statement

Assume ${A C}_{ω}$ . In a fixed-flow finite fence for a C² cooriented foliation, let cut paths $c_{t}, d_{t}$ be continuously defined for $a_{0} \leq t \leq r$ , with product the prescribed closed loop at each parameter. Suppose both factors are closed and leafwise null at $r$ . Extend their common closed/null interval downward maximally. Its lower endpoint $a$ has both factors closed. If the product at $a_{0}$ is essential, at least one factor at $a$ is essential; both are closed and null for $a < t \leq r$ .

Facts & Assumptions

Given: The fixed-flow fence, cut paths and parameters in the statement; the product at $a_{0}$ is essential for the essential-endpoint conclusion.

[F1]

Fixed-flow fences supply continuous endpoint tracks and a finite crossing word (Fixed transverse fences and their finite crossing words). Null caps persist locally under compact transverse deformation (A compact leafwise nullhomotopy persists under a transverse deformation), with prescribed boundaries realized by unique transverse roots (A fixed cap product glues by unique transverse flow roots).

[F2]

In the Hausdorff ambient manifold equality of continuous endpoint tracks is a closed condition: unequal limiting endpoints have disjoint neighborhoods. See also In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones.

[F3]

The choice hypothesis is The countable-choice principle used in the foliation pair.

Proof

technique · direct

1.1F1givenconstruct

Let $U$ consist of parameters where both factors are closed and null. At such a parameter a fixed null cap makes each cut-loop holonomy the identity on a neighborhood. Its endpoints run along the same fixed-flow track; uniqueness of plaque continuation and transverse-flow roots therefore keeps the prescribed cut path closed nearby. F1 then transports its compact cap. Doing this for both factors shows that $U$ is relatively open and contains $r$ . Nullity of the product alone would not suffice.

2.1F2step 1.1

Let $(a, r]$ be the component of $U$ below $r$ , including the initial endpoint if it belongs to $U$ . Continuity and F2 make both factors closed at $a$ . If $a > a_{0}$ and both were null there, step 1.1 would extend their common interval below $a$ , contradicting maximality. Thus one factor is essential at a proper lower endpoint.

3.1F1F3step 1.1step 2.1∎

If $a = a_{0}$ and the initial product is essential, both factors cannot be null there, since their product would then be null. At least one is essential in this case too. Both factors are closed and null for every $a < t \leq r$ , giving the selected essential endpoint its genuine one-sided null family. Only a fixed finite fence and one cap per factor are used.

Depends on

Used by

Compatible arbitrary pi fence reduction Lemma

Dependency tree · two levels

35 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

S. P. Novikov, The Topology of Foliations (complete English translation) (standard reference, not scraped)
Mark Brittenham, Foliations and the Topology of 3-Manifolds, classes 11-20 (standard reference, not scraped)