Fiber-homotopy equivalence

In algebraic topology, a fiber-homotopy equivalence is a map over a space B that has homotopy inverse over B (that is we require a homotopy be a map over B for each time t.) It is a relative analog of a homotopy equivalence between spaces.

Given maps p:DB, q:EB, if ƒ:DE is a fiber-homotopy equivalence, then for any b in B the restriction

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

Proposition  Let be fibrations. Then a map 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). We write 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 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 ; that is, .

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since , we have: . Since p is a fibration, the homotopy lifts to a homotopy from g to, say, g' that satisfies . 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:DD is over B via p and . Let be a homotopy from ƒ to . Then, since and since p is a fibration, the homotopy lifts to a homotopy ; explicitly, we have . Note also is over B.

We show is a left homotopy inverse of ƒ over B. Let be the homotopy given as the composition of homotopies . Then we can find a homotopy K from the homotopy pJ to the constant homotopy . Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:


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