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Boundary product function on a collared cobordism

Statement

Assume ${A C}_{ω}$ . Let $(W; M_{0}, M_{1})$ be a compact collared triad with fixed collar coordinates $t_{0} : M_{0} \times [0, 1) \to W$ and $t_{1} : M_{1} \times [0, 1) \to W$ having disjoint images; in function formulas, $t_{i}$ denotes the second coordinate of the inverse collar map. There is a smooth $h : W \to [0, 1]$ with $h^{- 1} (0) = M_{0}$ , $h^{- 1} (1) = M_{1}$ , $h = t_{0} / 3$ on a neighbourhood of $M_{0}$ , $h = 1 - t_{1} / 3$ on a neighbourhood of $M_{1}$ , $1 / 3 < h < 2 / 3$ outside the two collar neighbourhoods, and no critical point in a neighbourhood of $\partial W$ .

Facts & Assumptions

[F1]

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

[F2]

Collar neighborhood theorem: Assume ${A C}_{ω}$ . Every smooth manifold with boundary has a smooth collar.

[F3]

Smooth partitions of unity exist on manifolds with boundary: Assume ${A C}_{ω}$ . Every open cover of a smooth manifold with boundary admits a smooth partition of unity subordinate to it.

[F5]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): The Axiom of Countable Choice, written ${A C}_{ω}$ , is the following statement: for every family $(X_{n})_{n \in N}$ of nonempty sets indexed by $N$ there is a function $f$ with domain $N$ such that $f (n) \in X_{n}$ for every $n \in N$ .

[F6]

Smooth cobordism triad for Morse theory: A smooth cobordism triad $(W; M_{0}, M_{1})$ consists of a compact smooth $n$ -manifold with boundary $W$ , for $n \geq 1$ two closed embedded smooth $(n - 1)$ -submanifolds $M_{0}, M_{1} \subseteq \partial W$ with $\partial W = M_{0} ⊔ M_{1}$ , 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 $(W; M_{0}, M_{1})$ and the two fixed collars $t_{0}, t_{1}$ with disjoint images.

1.1F1F6algebra

Put $U_{0} : = t_{0} (M_{0} \times [0, 1 / 4))$ , $U_{1} : = t_{1} (M_{1} \times [0, 1 / 4))$ and $U_{2} : = W ∖ (t_{0} (M_{0} \times [0, 1 / 8]) \cup t_{1} (M_{1} \times [0, 1 / 8]))$ . Each $U_{i}$ is open, and they cover $W$ : a point outside the two closed strips lies in $U_{2}$ , while a point of $t_{0} (M_{0} \times [0, 1 / 8])$ lies in $U_{0}$ and a point of $t_{1} (M_{1} \times [0, 1 / 8])$ lies in $U_{1}$ .

2.1F3F5step 1.1construct

By [F3] choose a smooth partition of unity $(ψ_{0}, ψ_{1}, ψ_{2})$ subordinate to $(U_{0}, U_{1}, U_{2})$ ; choose a smooth scalar cutoff $b : [0, 1) \to [0, 1]$ equal to one for $t \leq 1 / 16$ and zero for $t \geq 1 / 8$ , obtained by integrating a nonnegative smooth bump in $(1 / 16, 1 / 8)$ and taking its normalized complementary integral. Define $β = b (t_{0})$ on the first collar and $γ = b (t_{1})$ on the second, extending both by zero off their collar images; their supports lie in $U_{0}, U_{1}$ and they are smooth up to the faces. Set $ϕ_{0} : = β + (1 - β) (1 - γ) ψ_{0}$ , $ϕ_{1} : = (1 - β) γ + (1 - β) (1 - γ) ψ_{1}$ and $ϕ_{2} : = (1 - β) (1 - γ) ψ_{2}$ . Then $ϕ_{0} + ϕ_{1} + ϕ_{2} = β + (1 - β) [(1 - γ) (ψ_{0} + ψ_{1} + ψ_{2}) + γ] = 1$ , each $ϕ_{i}$ is supported in $U_{i}$ , and $ϕ_{0} = 1$ on the neighbourhood ${β = 1}$ of $M_{0}$ while $ϕ_{1} = 1$ on the neighbourhood ${γ = 1}$ of $M_{1}$ , because $β$ vanishes on $U_{1}$ and $γ$ vanishes on $U_{0}$ .

3.1F2F3step 2.1construct

Define the functions $g_{0} : = t_{0} / 3$ on $U_{0}$ (where $t_{0} : M_{0} \times [0, 1 / 4) \to W$ is inverted on its image), $g_{1} : = 1 - t_{1} / 3$ on $U_{1}$ , and $g_{2} : = 1 / 2$ on $U_{2}$ , and set $h : = \sum_{i} ϕ_{i} g_{i}$ , a smooth function on $W$ with values in $[0, 1]$ because $0 \leq t_{0} / 3 < 1 / 12$ , $11 / 12 < 1 - t_{1} / 3 \leq 1$ and $1 / 2$ , and the $ϕ_{i}$ form a partition of unity.

4.1step 2.1step 3.1algebra

Values at the faces: near $M_{0}$ one has $ϕ_{0} = 1$ and $ϕ_{1} = ϕ_{2} = 0$ , so $h = t_{0} / 3$ , which vanishes exactly on $M_{0}$ and is positive elsewhere; near $M_{1}$ one has $ϕ_{1} = 1$ , so $h = 1 - t_{1} / 3$ , which equals $1$ exactly on $M_{1}$ . At every interior point all active collar values are strictly between zero and one, as is $g_{2} = 1 / 2$ ; their convex combination is therefore strictly between zero and one. Hence the first equality gives $h^{- 1} (0) = M_{0}$ and the second gives $h^{- 1} (1) = M_{1}$ .

4.2step 1.1step 3.1algebra

Outside the two collar neighbourhoods $U_{0} \cup U_{1}$ only the term with $ϕ_{2}$ contributes, so $h = 1 / 2$ there; in particular $1 / 3 < h < 2 / 3$ on $W ∖ (U_{0} \cup U_{1})$ .

5.1step 4.1algebra∎

No critical point occurs in a neighbourhood of $\partial W$ : in the coordinates $(p, t) \in M_{0} \times [0, 1 / 16)$ near $M_{0}$ one has $h = t / 3$ , whose differential is $\frac{1}{3} d t \neq 0$ , so $d h$ has no zero there; the same computation with $h = 1 - t / 3$ gives the statement near $M_{1}$ . These two open collar strips give a neighbourhood of $\partial W = M_{0} ⊔ M_{1}$ that is free of critical points of $h$ , as asserted.

Boundary product function on a collared cobordism

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources