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Boundary product function on a collared cobordism

Statement

Assume ACω. Let (W;M0,M1) be a compact collared triad with fixed collar coordinates t0:M0×[0,1)→W and t1:M1×[0,1)→W having disjoint images; in function formulas, ti denotes the second coordinate of the inverse collar map. There is a smooth h:W→[0,1] with h−1(0)=M0, h−1(1)=M1, h=t0/3 on a neighbourhood of M0, h=1−t1/3 on a neighbourhood of M1, 1/3<h<2/3 outside the two collar neighbourhoods, and no critical point in a neighbourhood of ∂W.

Facts & Assumptions

[F1]

Smooth collars of a manifold boundary: A smooth collar is a smooth embedding c:∂M×[0,ε)→M such that c(p,0)=p and whose image is an open neighbourhood of ∂M in M.

[F2]

Collar neighborhood theorem: Assume ACω. Every smooth manifold with boundary has a smooth collar.

[F3]

Smooth partitions of unity exist on manifolds with boundary: Assume ACω. Every open cover of a smooth manifold with boundary admits a smooth partition of unity subordinate to it.

[F5]

The Axiom of Countable Choice (ACω): The Axiom of Countable Choice, written ACω, is the following statement: for every family (Xn)n∈N of nonempty sets indexed by N there is a function f with domain N such that f(n)∈Xn for every n∈N.

[F6]

Smooth cobordism triad for Morse theory: A smooth cobordism triad (W;M0,M1) consists of a compact smooth n-manifold with boundary W, for n≥1 two closed embedded smooth (n−1)-submanifolds M0,M1⊆∂W with ∂W=M0⊔M1, and fixed collars of both faces in W. For n=0, both faces and collar domains are empty; no manifold of dimension −1 is introduced.

Proof

Given: The triad (W;M0,M1) and the two fixed collars t0,t1 with disjoint images.

1.1F1F6algebra

Put U0:=t0(M0×[0,1/4)), U1:=t1(M1×[0,1/4)) and U2:=W∖(t0(M0×[0,1/8])∪t1(M1×[0,1/8])). Each Ui is open, and they cover W: a point outside the two closed strips lies in U2, while a point of t0(M0×[0,1/8]) lies in U0 and a point of t1(M1×[0,1/8]) lies in U1.

2.1F3F5step 1.1construct

By [F3] choose a smooth partition of unity (ψ0,ψ1,ψ2) subordinate to (U0,U1,U2); choose a smooth scalar cutoff b:[0,1)→[0,1] equal to one for t≤1/16 and zero for t≥1/8, obtained by integrating a nonnegative smooth bump in (1/16,1/8) and taking its normalized complementary integral. Define β=b(t0) on the first collar and γ=b(t1) on the second, extending both by zero off their collar images; their supports lie in U0,U1 and they are smooth up to the faces. Set ϕ0:=β+(1−β)(1−γ)ψ0, ϕ1:=(1−β)γ+(1−β)(1−γ)ψ1 and ϕ2:=(1−β)(1−γ)ψ2. Then ϕ0+ϕ1+ϕ2=β+(1−β)[(1−γ)(ψ0+ψ1+ψ2)+γ]=1, each ϕi is supported in Ui, and ϕ0=1 on the neighbourhood {β=1} of M0 while ϕ1=1 on the neighbourhood {γ=1} of M1, because β vanishes on U1 and γ vanishes on U0.

3.1F2F3step 2.1construct

Define the functions g0:=t0/3 on U0 (where t0:M0×[0,1/4)→W is inverted on its image), g1:=1−t1/3 on U1, and g2:=1/2 on U2, and set h:=∑iϕigi, a smooth function on W with values in [0,1] because 0≤t0/3<1/12, 11/12<1−t1/3≤1 and 1/2, and the ϕi form a partition of unity.

4.1step 2.1step 3.1algebra

Values at the faces: near M0 one has ϕ0=1 and ϕ1=ϕ2=0, so h=t0/3, which vanishes exactly on M0 and is positive elsewhere; near M1 one has ϕ1=1, so h=1−t1/3, which equals 1 exactly on M1. At every interior point all active collar values are strictly between zero and one, as is g2=1/2; their convex combination is therefore strictly between zero and one. Hence the first equality gives h−1(0)=M0 and the second gives h−1(1)=M1.

4.2step 1.1step 3.1algebra

Outside the two collar neighbourhoods U0∪U1 only the term with ϕ2 contributes, so h=1/2 there; in particular 1/3<h<2/3 on W∖(U0∪U1).

5.1step 4.1algebra∎

No critical point occurs in a neighbourhood of ∂W: in the coordinates (p,t)∈M0×[0,1/16) near M0 one has h=t/3, whose differential is 13dt≠0, so dh has no zero there; the same computation with h=1−t/3 gives the statement near M1. These two open collar strips give a neighbourhood of ∂W=M0⊔M1 that is free of critical points of h, as asserted.

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