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Compact transverse complementary intersections are finite

Statement

Let f:Xx→Mn be smooth with X compact, let Z⊆M be a closed embedded submanifold, and suppose f is transverse to Z with x+dim⁡Z=n. Then f−1(Z) is finite (possibly empty), so #f−1(Z) is a well-defined nonnegative integer. Likewise, if Aa,Bb⊆M are transverse embedded submanifolds of M with a+b=n and one of A,B is compact while the other is closed, then A∩B is finite. Both compactness of the relevant source and closedness of the other factor are used: a zero-dimensional manifold is discrete, and a compact discrete space is finite.

Facts & Assumptions

Given: A smooth map f:Xx→Mn with X compact, Z⊆M a closed embedded submanifold, f⋔Z and x+dim⁡Z=n; and the corresponding submanifold situation.

[F1]

Under these hypotheses f−1(Z) is identified with X×MZ by p↦(p,f(p)); projection to X is its inverse. It is a 0-dimensional embedded submanifold of X by The transverse preimage theorem; in a slice chart with k=0 each point is an isolated point of f−1(Z), with the subspace topology (Transverse complementary-dimensional intersection sets, Embedded submanifolds and slice charts, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F2]

Preimages of closed sets under continuous maps are closed; the points of X with f(x)∈Z form the preimage of the closed set Z (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and f(A‾)⊆f(A)‾).

[F4]

A compact discrete topological space is finite: its singleton open cover has a finite subcover, whose union is a finite set equal to the whole space; the discrete topology on an infinite set is not compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

Proof

technique · direct, from discreteness and compactness
1.1F1given

By [F1] the set f−1(Z) is a 0-dimensional embedded submanifold of X, hence discrete in its subspace topology: each of its points has a slice chart in which it is the only point of the set in that chart.

2.1F2F3F4step 1.1given

The set f−1(Z) is the preimage of the closed set Z under the continuous map f, since smooth maps are continuous, hence closed in X by [F2], hence compact by compactness of X and [F3]. A compact discrete space is finite by [F4], so #f−1(Z) is a well-defined nonnegative integer; the empty case is included.

3.1step 2.1givenalgebra∎

For the submanifold case, apply the map case to the inclusion of the compact factor: if A is compact and B is closed in M, then the inclusion iA:A→M is smooth with A compact, iA⋔B because A⋔B, and iA−1(B)=A∩B, so 2.1 gives that A∩B is finite; the roles of A and B may be exchanged. No choice axiom is used.

Depends on

Used by

Dependency tree · two levels

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Sources