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A nontransverse pullback need not reproduce the rank of a foliation

Statement refuted

Assume Countable Choice ACω (The Axiom of Countable Choice (ACω)). The following inference is false: for an arbitrary smooth map f:N→M and a regular foliation F of M, the inverse-image spaces Dt∗:=(dft)−1(Tf(t)F) form a smooth distribution of constant rank dim⁡N−codim⁡F and define a regular pullback foliation whose leaves are the connected components of leaf preimages.

Counterexample. Let M=R2 with the horizontal foliation, let N=R, and let f(t)=(0,t2). Since dft(v)=(0,2tv) and Tf(t)F=R×{0}, one has Dt∗={0} for t≠0 but D0∗=T0R. Thus the rank jumps from 0 to 1 at 0, so D∗ is not a regular distribution. The map is transverse to F for t≠0 and nontransverse at 0. This refutes the rank and regular-distribution inference without the transversality hypothesis.

Facts & Assumptions

Given: The horizontal foliation F of R2, the map f:R→R2, f(t)=(0,t2), and the family of inverse-image spaces Dt∗=(dft)−1(Tf(t)F).

[F1]

The horizontal foliation F of R2 is the regular foliation whose leaves are the lines R×{c}; its tangent distribution is Tf(t)F=R×{0} at every point, a rank-one subbundle of TR2 (Flat charts for a distribution, Integrable distributions).

[F2]

A smooth map f is transverse to F at t exactly when dft(TtN)+Tf(t)F=Tf(t)M, and the inverse-image convention for the pullback is Dt∗=(dft)−1(Tf(t)F) (Smooth maps transverse to a regular foliation, The pullback foliation under a transverse map).

[F3]

For smooth f the differential dft is the linear map of tangent spaces induced by f, computed in coordinates by the Jacobian matrix (The differential of a smooth map).

[F4]

A smooth distribution of rank k assigns to every point a k-dimensional subspace as a smooth vector subbundle, so its rank is constant; the linear preimage of a linear subspace under a linear map is a linear subspace (Vector subbundles).

Counterexample

1.1F3given

In coordinates on R2 and R the map f(t)=(0,t2) has Jacobian (0,2t)T, so dft(v)=(0,2tv) for every v∈TtR, by [F3].

2.1F1F2step 1.1algebra

Hence dft(v)∈Tf(t)F=R×{0} holds exactly when 2tv=0. Therefore Dt∗={v:2tv=0} equals {0} for t≠0 and equals T0R at t=0.

2.2F1F2step 1.1

The map is transverse to F for t≠0: there dft(TtR) is the vertical line {0}×R, which together with Tf(t)F=R×{0} spans Tf(t)R2. At t=0 the differential vanishes, so df0(T0R)={0} and the sum df0(T0R)+Tf(0)F=R×{0} is a proper subspace of Tf(0)R2: the map is not transverse at 0. Thus the failure of the rank conclusion occurs exactly at the point where transversality fails, and the transversality hypothesis of The pullback foliation under a transverse map is essential.

3.1F4step 2.1

The rank of Dt∗ is 0 for t≠0 and 1 at t=0; in particular D∗ is not a smooth distribution of constant rank dim⁡N−codim⁡F=1−1=0, and it is not a vector subbundle of TR near 0. So the first two conclusions of the inference fail.

4.1F1F2step 2.1step 3.1∎

Finally the horizontal leaf Lc=R×{c} has preimage f−1(Lc)={t:t2=c}, a finite set or empty; its connected components are points, whose tangent spaces are {0}, while D0∗=T0R. So the leaf-preimage description is not compatible with the inverse-image spaces at 0 either, and the inference is false in every one of its clauses. The claim is refuted.

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