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: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.

Proposition — Let $p: D \to B, q: E \to B$ be fibrations. Then a map $f: D \to E$ 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). 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: