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

[math]\displaystyle{ f: p^{-1}(b) \to q^{-1}(b) }[/math]

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

Proposition — Let [math]\displaystyle{ p: D \to B, q: E \to B }[/math] be fibrations. Then a map [math]\displaystyle{ f: D \to E }[/math] 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 1999). We write [math]\displaystyle{ \sim_B }[/math] 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 [math]\displaystyle{ g f \sim_{B} \operatorname{id} }[/math] 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 [math]\displaystyle{ h \sim f }[/math]; that is, [math]\displaystyle{ f g \sim_{B} \operatorname{id} }[/math].

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since [math]\displaystyle{ fg \sim \operatorname{id} }[/math], we have: [math]\displaystyle{ pg = qfg \sim q }[/math]. Since p is a fibration, the homotopy [math]\displaystyle{ pg \sim q }[/math] lifts to a homotopy from g to, say, g' that satisfies [math]\displaystyle{ pg' = q }[/math]. 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 [math]\displaystyle{ f \sim \operatorname{id}_D }[/math]. Let [math]\displaystyle{ h_t }[/math] be a homotopy from ƒ to [math]\displaystyle{ \operatorname{id}_D }[/math]. Then, since [math]\displaystyle{ p h_0 = p }[/math] and since p is a fibration, the homotopy [math]\displaystyle{ ph_t }[/math] lifts to a homotopy [math]\displaystyle{ k_t: \operatorname{id}_D \sim k_1 }[/math]; explicitly, we have [math]\displaystyle{ p h_t = p k_t }[/math]. Note also [math]\displaystyle{ k_1 }[/math] is over B.

We show [math]\displaystyle{ k_1 }[/math] is a left homotopy inverse of ƒ over B. Let [math]\displaystyle{ J: k_1 f \sim h_1 = \operatorname{id}_D }[/math] be the homotopy given as the composition of homotopies [math]\displaystyle{ k_1 f \sim f = h_0 \sim \operatorname{id}_D }[/math]. Then we can find a homotopy K from the homotopy pJ to the constant homotopy [math]\displaystyle{ p k_1 = p h_1 }[/math]. Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:

[math]\displaystyle{ 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}. }[/math]

References

• May, J. Peter (1999). A concise course in algebraic topology. Chicago Lectures in Mathematics. Chicago: University of Chicago Press. ISBN 0-226-51182-0. OCLC 41266205. http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf. (See chapter 6.)