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Can someone give me a hint on proving that the fibre of a fibration $f: Y \to X$ is homotopy equivalent to its homotopy fibre $Y \times_X X^I$?

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Not a hint, but a couple of arguments and references can be found in this MO-thread. –  t.b. May 10 '11 at 11:55
    
Thanks, I already found this. But I don't want the complete argument. –  user5262 May 10 '11 at 12:33
    
The first step would be to construct a continuous function from one to the other. There appears to be a good choice from the fibre to the homotopy fibre. –  Ryan Budney May 10 '11 at 17:38

2 Answers 2

The homotopy fiber of a map is only defined up to homotopy equivalence. I think this is a crucial point to your question. One model for it is given by the following construction.

Let $f: X \to Y$ be any map. Then we may factor it as a homotopy equivalence followed by a fibration, ie $X \to Z \to Y$, call the last map $g: Z \to Y$ which is a fibration. Now one model for the homotopy fiber of $f$ is just the honest fiber of the map $g$. In the case that $f$ itself is a fibration one may choose the obvious factorization to see that its fiber is one model for a homotopy fiber.

But since you asked for this very explicit model, at least assuming that all spaces are CW complexes, here is a brief idea of an argument. Comparing the long exact sequences of the fibration $f$ and the long exact sequence associated to the homotopy fiber of $f$ together with the "good choice" of a map from the fiber to the homotopy fiber and a 5 Lemma argument shows that fiber and homotopy fiber are weakly equivalent. By Whitehead then they are also homotopy equivalent. By the good choice I mean the following: A map from the fiber $F$ to $Y\times_{X} X^{I}$ is given by compatible maps $F \to Y$ which you choose to be the inclusion and $F to X^{I}$ which can be chosen as the constant map at the basepoint of $X$.

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This is a consequence of the main theorem in

R. Brown and P.R. Heath, "Coglueing homotopy equivalences", Math.Z. 113 (1970) 313-325,

available here, which gives conditions for a pullback of homotopy equivalences to be a homotopy equivalence. A fibre is the pullback over a point. One uses the fact that any map, which in this instance is a fibration, factors through a homotopy equivalence and a fibration, as given by mland. The theorem is the dual of the gluing theorem for homotopy equivalences, which can be found in Topology and Groupoids, and has a dual proof. This proof gives also good control over the homotopies involved.

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313-215 can be right. –  Rasmus Jul 19 '13 at 13:32

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