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Fiber-homotopy equivalence

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In algebraic topology, a fiber-homotopy equivalence is a map over a space B that has homotopy inverse over B (that is if h t {\displaystyle h_{t}} is a homotopy between the two maps, h t {\displaystyle h_{t}} is a map over B for t.) It is a relative analog of a homotopy equivalence between spaces.

Given maps p: D → B, q: E → B, if ƒ: D → E is a fiber-homotopy equivalence, then for any b in B the restriction

f : p − 1 ( b ) → q − 1 ( b ) {\displaystyle f:p^{-1}(b)\to q^{-1}(b)}

is a homotopy equivalence. If p, q are fibrations, this is always the case for homotopy equivalences by the next proposition.

Proposition — Let p : D → B , q : E → B {\displaystyle p:D\to B,q:E\to B} be fibrations. Then a map f : D → E {\displaystyle f:D\to E} over B is a homotopy equivalence if and only if it is a fiber-homotopy equivalence.

Proof of the proposition

The following proof is based on the proof of Proposition in Ch. 6, § 5 of (May 1999). We write ∼ B {\displaystyle \sim _{B}} for a homotopy over B.

We first note that it is enough to show that ƒ admits a left homotopy inverse over B. Indeed, if g f ∼ B id {\displaystyle gf\sim _{B}\operatorname {id} } with g over B, then g is in particular a homotopy equivalence. Thus, g also admits a left homotopy inverse h over B and then formally we have h ∼ f {\displaystyle h\sim f} ; that is, f g ∼ B id {\displaystyle fg\sim _{B}\operatorname {id} } .

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since f g ∼ id {\displaystyle fg\sim \operatorname {id} } , we have: p g = q f g ∼ q {\displaystyle pg=qfg\sim q} . Since p is a fibration, the homotopy p g ∼ q {\displaystyle pg\sim q} lifts to a homotopy from g to, say, g' that satisfies p g ′ = q {\displaystyle pg'=q} . Thus, we can assume g is over B. Then it suffices to show gƒ, which is now over B, has a left homotopy inverse over B since that would imply that ƒ has such a left inverse.

Therefore, the proof reduces to the situation where ƒ: D → D is over B via p and f ∼ id D {\displaystyle f\sim \operatorname {id} _{D}} . Let h t {\displaystyle h_{t}} be a homotopy from ƒ to id D {\displaystyle \operatorname {id} _{D}} . Then, since p h 0 = p {\displaystyle ph_{0}=p} and since p is a fibration, the homotopy p h t {\displaystyle ph_{t}} lifts to a homotopy k t : id D ∼ k 1 {\displaystyle k_{t}:\operatorname {id} _{D}\sim k_{1}} ; explicitly, we have p h t = p k t {\displaystyle ph_{t}=pk_{t}} . Note also k 1 {\displaystyle k_{1}} is over B.

We show k 1 {\displaystyle k_{1}} is a left homotopy inverse of ƒ over B. Let J : k 1 f ∼ h 1 = id D {\displaystyle J:k_{1}f\sim h_{1}=\operatorname {id} _{D}} be the homotopy given as the composition of homotopies k 1 f ∼ f = h 0 ∼ id D {\displaystyle k_{1}f\sim f=h_{0}\sim \operatorname {id} _{D}} . Then we can find a homotopy K from the homotopy pJ to the constant homotopy p k 1 = p h 1 {\displaystyle pk_{1}=ph_{1}} . Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:

k 1 f = J 0 = L 0 , 0 ∼ B L 0 , 1 ∼ B L 1 , 1 ∼ B L 1 , 0 = J 1 = id . {\displaystyle k_{1}f=J_{0}=L_{0,0}\sim _{B}L_{0,1}\sim _{B}L_{1,1}\sim _{B}L_{1,0}=J_{1}=\operatorname {id} .}

References

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