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Lifts of circle-loop concatenations and reversals

Statement

Let α and β be based loops in R/Z, and let their lifts from zero be α~ and β~, with terminal values m and n. The lift of αβ from zero is

αβ~(t)={α~(2t),0t1/2,m+β~(2t1),1/2t1,

and it ends at m+n. The lift of the reversed loop αˉ from zero is

αˉ~(t)=α~(1t)m,

and it ends at m. Thus lifts of circle-loop concatenations and reversals have endpoints equal to the sum and the negative of the original endpoints.

Facts & Assumptions

Given: Based loops α,β, their lifts α~,β~ from zero, and terminal values m=α~(1) and n=β~(1).

[L1]

For a based circle loop γ with lift γ~ from zero, the terminal value γ~(1) is an integer and deg(γ)=γ~(1) (The degree of a based circle loop).

[L2]

The product [α][β] traverses α first and β second, and is represented by αβ (Based loops and the fundamental group).

[L3]

A path through a covering has a unique lift once its initial point is prescribed (Existence and uniqueness of path lifts through a covering map).

[L4]

Functions continuous on each member of a finite closed cover, and agreeing where the pieces meet, paste to a continuous function (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L5]

For the quotient projection p, one has p(x+k)=p(x) for every real x and integer k (The circle as S1=R/Z with basepoint [0]).

Proof

technique · direct
1.1

Define γ by the displayed two-piece formula. At t=1/2 the left value is α~(1)=m and the right value is m+β~(0)=m, so [L4] and [L6] make γ continuous. It starts at zero. By [L5], its first half projects to α(2t) and its second half to β(2t1), in the order fixed by [L2], so pγ=αβ; its endpoint is m+n.

L1L2L4L5L6
1.2

Define δ(t)=α~(1t)m. It is continuous by [L6], begins at mm=0, and ends at 0m=m. Since mZ by [L1], [L5] gives p(δ(t))=p(α~(1t))=α(1t)=αˉ(t).

L1L5L6algebra
2.1

Both γ and δ are lifts with initial point zero, so uniqueness in [L3] identifies them with the defining lifts of αβ and αˉ. Their endpoints are therefore m+n and m, respectively.

step 1.1step 1.2L3

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 106 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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