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Normalized comparison isomorphisms are transitive

Statement

Let t,u,w be signed words representing the same braid. Then γu,w∘γt,u=γt,win Hom⁡Kb(F(t),F(w))=Q⋅[γt,w]. In particular γt,t=id and γu,tγt,u=id, so the maps γ form a transitive system of homotopy equivalences between the word complexes of a fixed braid; consequently the multiplication comparisons of the next theorem are well defined on chosen representatives.

Facts & Assumptions

Given: Signed words t,u,w with the same product, the word complexes F(t),F(u),F(w) of The Rouquier complex of a braid word, and the normalized maps γt,u,γu,w,γt,w of Derived comparisons give unique normalized homotopy maps.

[F1]

Uniqueness. For words a,b with the same product, Hom⁡Kb(F(a),F(b)) is one-dimensional in internal degree 0, the localization map to Hom⁡Db is an isomorphism, and γa,b is the unique homotopy class whose derived image is the comparison ca,b of the graph models. (Derived comparisons give unique normalized homotopy maps)

[F2]

Transitivity of the comparisons. The derived comparisons satisfy cu,wct,u=ct,w and ct,t=id; they are the multiplication isomorphisms of the words through the standard graph bimodules. (Canonical comparisons between standard graph tensor products)

Proof

technique · direct
1.1F1

The composite γu,w∘γt,u is an element of Hom⁡Kb(F(t),F(w)), which by [F1] is one-dimensional in internal degree 0; its derived image is cu,wct,u because localization is a functor on the homotopy categories in which the γ become isomorphisms.

2.1F1F2step 1.1

By [F2] cu,wct,u=ct,w, which is the derived image of γt,w by [F1]; two elements of the one-dimensional space with the same nonzero derived image coincide, so γu,wγt,u=γt,w.

3.1F1F2step 2.1∎

Taking u=t=w and using ct,t=id gives γt,t=id by the same uniqueness argument; then γt,u and γu,t are mutually inverse homotopy equivalences because both composites equal the corresponding identity maps, and the identity is the unique degree-zero endomorphism class whose derived image is the normalized identity comparison.

Depends on

Used by

Dependency tree · two levels

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Sources