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

${\displaystyle f:p^{-1}(b)\to q^{-1}(b)}$

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

Proposition — Let ${\displaystyle p:D\to B,q:E\to B}$ be fibrations. Then a map ${\displaystyle f:D\to E}$ 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) harv error: no target: CITEREFMay (help). We write ${\displaystyle \sim _{B}}$ 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 ${\displaystyle gf\sim _{B}\operatorname {id} }$ 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 ${\displaystyle h\sim f}$; that is, ${\displaystyle fg\sim _{B}\operatorname {id} }$.

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since ${\displaystyle fg\sim \operatorname {id} }$, we have: ${\displaystyle pg=qfg\sim q}$. Since p is a fibration, the homotopy ${\displaystyle pg\sim q}$ lifts to a homotopy from g to, say, g' that satisfies ${\displaystyle pg'=q}$. 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 ${\displaystyle f\sim \operatorname {id} _{D}}$. Let ${\displaystyle h_{t}}$ be a homotopy from ƒ to ${\displaystyle \operatorname {id} _{D}}$. Then, since ${\displaystyle ph_{0}=p}$ and since p is a fibration, the homotopy ${\displaystyle ph_{t}}$ lifts to a homotopy ${\displaystyle k_{t}:\operatorname {id} _{D}\sim k_{1}}$; explicitly, we have ${\displaystyle ph_{t}=pk_{t}}$. Note also ${\displaystyle k_{1}}$ is over B.

We show ${\displaystyle k_{1}}$ is a left homotopy inverse of ƒ over B. Let ${\displaystyle J:k_{1}f\sim h_{1}=\operatorname {id} _{D}}$ be the homotopy given as the composition of homotopies ${\displaystyle k_{1}f\sim f=h_{0}\sim \operatorname {id} _{D}}$. Then we can find a homotopy K from the homotopy pJ to the constant homotopy ${\displaystyle pk_{1}=ph_{1}}$. Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:

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

References

• May, J.P. A Concise Course in Algebraic Topology, (1999) Chicago Lectures in Mathematics ISBN 0-226-51183-9 (See chapter 6.)