Fiber-homotopy equivalence

In algebraic topology, a fiber-homotopy equivalence is a homotopy equivalence between fibers of maps into a space B from spaces D and E (that is, a map between preimages that is bidirectionally invertible up to homotopy). It is a fiber-wise 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
 * $$f: p^{-1}(b) \to q^{-1}(b)$$

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

Proof of the proposition
The following proof is based on the proof of Proposition in Ch. 6, § 5 of. We write $$\sim_B$$ 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 $$g f \sim_{B} \operatorname{id}$$ 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 $$h \sim f$$; that is, $$f g \sim_{B} \operatorname{id}$$.

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since $$fg \sim \operatorname{id}$$, we have: $$pg = qfg \sim q$$. Since p is a fibration, the homotopy $$pg \sim q$$ lifts to a homotopy from g to, say, g '  that satisfies $$pg' = q$$. 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 $$f \sim \operatorname{id}_D$$. Let $$h_t$$ be a homotopy from ƒ to $$\operatorname{id}_D$$. Then, since $$p h_0 = p$$ and since p is a fibration, the homotopy $$ph_t$$ lifts to a homotopy $$k_t: \operatorname{id}_D \sim k_1$$; 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 $$J: k_1 f \sim h_1 = \operatorname{id}_D$$ be the homotopy given as the composition of homotopies $$k_1 f \sim f = h_0 \sim \operatorname{id}_D$$. Then we can find a homotopy K from the homotopy pJ to the constant homotopy $$p k_1 = p h_1$$. Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:
 * $$k_1 f = J_0 = L_{0, 0} \sim_B L_{0, 1} \sim_B L_{1, 1} \sim_B L_{1, 0} = J_1 = \operatorname{id}.$$