In algebraic topology, a fiber-homotopy equivalence is a map over a space B that has homotopy inverse over B (that is if is a homotopy between the two maps, is a map over B for t.) It is a relative analog of a homotopy equivalence between spaces.
Given maps p: D â B, q: E â B, if ÃÂ: D â E 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.
There is also a completely analogous notion of a cofiber-homotopy equivalence, a homotopy equivalence such that the homotopies connecting the compositions and the identities are homotopies under the base. Then, similar to the above, we have:
The following proof is based on the proof of Proposition in Ch. 6, ç 5 of . We write for a homotopy over B.
We first note that it is enough to show that ÃÂ 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 ÃÂ 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<nowiki>'</nowiki> that satisfies . Thus, we can assume g is over B. Then it suffices to show gÃÂ, which is now over B, has a left homotopy inverse over B since that would imply that ÃÂ has such a left inverse.
Therefore, the proof reduces to the situation where ÃÂ: D â D is over B via p and . Let be a homotopy from àto . 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 ÃÂ 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: