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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-01
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A smooth exhaustion separates the locally finite chart bands

Statement

Let Mn be a noncompact smooth manifold. Then there exist a smooth proper function ρ:MR, compact bands Km:=ρ1([m1,m+2])(m1), and smooth maps Hm:MRQm such that:

  1. each Hm is supported in a neighbourhood of Km;
  2. the supports of Hm and Hm are disjoint whenever mm(mod4) and mm;
  3. Hm separates points and tangent vectors on Km; and
  4. Hm2m everywhere.

Facts & Assumptions

Given: A noncompact smooth n-manifold M.

[L1]

The manifold admits a smooth proper exhaustion function ρ:MR (Every smooth manifold admits a smooth proper exhaustion function).

[L2]

A closed set inside an open set admits a smooth cutoff equal to 1 near the closed set and supported in the open set (A smooth Urysohn lemma for a closed set in an open set).

[F1]

A smooth manifold comes with smooth coordinate charts (Smooth manifolds and their smooth charts).

[L3]

Smooth maps that agree on overlaps paste over an open cover (Smooth maps paste over an open cover).

Proof

technique · direct
1.1

Choose a nonnegative smooth proper exhaustion ρ from [L1]. Let Km:=ρ1([m1,m+2])(m1). Each Km is compact and the family (Km) covers M. If mm(mod4) and mm, the defining intervals are separated by a positive gap.

L1givenconstruct
2.1

Put Om:=ρ1((m5/4,m+9/4)). Then KmOm, and OmOm= for distinct congruent indices modulo 4. If Km=, take Qm=1 and Hm=0; all four requirements for this index are then immediate. Henceforth suppose Km.

step 1.1construct
3.1

For every pKm, a chart from [F1] can be shrunk over a Euclidean ball to a coordinate domain (U,x) with pU, compact closure, and UOm. Applying [L2] to {p}U gives a smooth ϕ:M[0,1] supported in U and equal to 1 on an open neighbourhood V of p. The collection of all plateau neighbourhoods V obtainable in this way covers Km, so compactness selects finitely many data (Umj,xmj,ϕmj,Vmj), 1jrm, whose Vmj cover Km.

F1L2step 2.1choose
4.1

For each selected datum define a global block Bmj:MRn+1 by Bmj(q):={(ϕmj(q),ϕmj(q)xmj(q)),qUmj,0,qUmj. On the open cover Umj(Msupp(ϕmj)) the two formulas are smooth and agree on the overlap, so [L3] makes Bmj smooth. Set Hm:=(Bm1,,Bmrm):MRrm(n+1). Its support lies in the finite union of the compact sets supp(ϕmj)Om.

F1L3step 3.1construct
5.1

The map Hm separates points of Km: if Hm(p)=Hm(q), choose j with pVmj. Equality of the first coordinate of the jth block gives ϕmj(q)=1, and equality of the remaining coordinates gives xmj(p)=xmj(q), whence p=q. It also separates tangent vectors: for pVmj, the function ϕmj is locally constant with value 1, so the last n components of dBmj,p are dxmj,p, an isomorphism. Thus dHm,p is injective for every pKm.

step 3.1step 4.1algebra
6.1

Compact support makes Hm bounded. Choose Cm1 with Hm(q)Cm for every qM, and put Hm:=2mCm1Hm. This positive rescaling preserves support and both separation properties, and it gives Hm2m everywhere.

step 4.1step 5.1chooseconstruct
7.1

For a nonempty band, steps 4.1 and 6.1 put supp(Hm) inside Om; for an empty band, step 2.1 gives empty support. The sets Om are disjoint for distinct congruent indices modulo 4, so the corresponding supports are disjoint. Together with steps 1.1, 5.1, and 6.1, this proves all four stated properties.

step 1.1step 2.1step 4.1step 5.1step 6.1

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