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 if $h_t$ is a homotopy between the two maps, $h_t$ is a map over B for 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.

There is also a completely analogous notion of a cofiber-homotopy equivalence, a homotopy equivalence such that the homotopies connecting the compositions and the identities are homotopies under the base. Then, similar to the above, we have:

Proposition —  Let $i:A\to X,j:A\to Y$ be cofibrations. Then a map $f:X\to Y$ under A is a homotopy equivalence if and only if it is a cofiber-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 $gf{\sim }_B\mathrm{id}$ 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, $fg{\sim }_B\mathrm{id}$.

Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since $fg\sim \mathrm{id}$, 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 $p{g}^{\prime }=q$. 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 $f\sim {\mathrm{id}}_D$. Let $h_t$ be a homotopy from ƒ to ${\mathrm{id}}_D$. Then, since $p{h}_0=p$ and since p is a fibration, the homotopy $p{h}_t$ lifts to a homotopy $k_t:{\mathrm{id}}_D\sim k_1$; 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 $J:k_{1}f\sim h_1={\mathrm{id}}_D$ be the homotopy given as the composition of homotopies $k_{1}f\sim f=h_0\sim {\mathrm{id}}_D$. 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:

$k_{1}f=J_0=L_{0,0}{\sim }_B{L}_{0,1}{\sim }_B{L}_{1,1}{\sim }_B{L}_{1,0}=J_1=\mathrm{id}.$

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.) 
• Hatcher, Allen. "Algebraic topology". https://pi.math.cornell.edu/~hatcher/AT/ATpage.html.