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.

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 1999). 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 $gf\sim _{B}\operatorname {id}$ 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, $fg\sim _{B}\operatorname {id}$.

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 $ph_{0}=p$ 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 $ph_{t}=pk_{t}$. 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 $pk_{1}=ph_{1}$. Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:

$k_{1}f=J_{0}=L_{0,0}\sim _{B}L_{0,1}\sim _{B}L_{1,1}\sim _{B}L_{1,0}=J_{1}=\operatorname {id} .$

References

• May, J. Peter (1999). A concise course in algebraic topology (PDF). Chicago Lectures in Mathematics. Chicago: University of Chicago Press. ISBN 0-226-51182-0. OCLC 41266205. (See chapter 6.){{cite book}}: CS1 maint: postscript (link)