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: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 ) is a homotopy equivalence. If p, q are fibrations, this is always the case for homotopy equivalences by the next proposition.
The following proof is based on the proof of Proposition in Ch. 6, § 5 of (May). We write ∼ 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 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 ; that is, f g ∼ B id .
Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since f g ∼ id , we have: p g = q f g ∼ q . Since p is a fibration, the homotopy p g ∼ q lifts to a homotopy from g to, say, g' that satisfies p g ′ = 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 . Let h t be a homotopy from ƒ to id D . Then, since p h 0 = p and since p is a fibration, the homotopy p h t lifts to a homotopy k t : id D ∼ k 1 ; explicitly, we have p h t = p k t . Note also k 1 is over B.
We show k 1 is a left homotopy inverse of ƒ over B. Let J : k 1 f ∼ h 1 = id D be the homotopy given as the composition of homotopies k 1 f ∼ f = h 0 ∼ id D . Then we can find a homotopy K from the homotopy pJ to the constant homotopy p k 1 = p h 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 . 