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Let $p:E \to B$ be a fibration with fibre $F$ (denote the fibration $\xi$). Let us assume that the fibre is $(n-1)$ connected. There is a fundamental class $\iota_F \in H^n(F;\pi_n(F))$. We can define the characteristic class $\chi(\xi)$ of the fibre space $\xi$ as the cohomology class $\tau\left(\iota_F\right) \in H^{n+1}(B;\pi_n(F))$ where $\tau$ is the transgression.

The question is, as per the title, to show that the characteristic class of a fibration is a fibre homtopy invariant.

Two fibrations $\xi=(E,p,B;F)$ and $\xi'=(E',p',B;F')$ are said to be fibre homotopy equivalent if there is a homotpy equivalence $\phi:E \to E'$ and $\psi:E' \to E$ such that $\psi \phi = 1_{E'}$ and $\phi \psi = 1_{E}$, and these homotpy equivalences are compatible with the projections.

I can see that if we have fibre homotopy equivalent spaces, then the restriction of $\psi$ and $\phi$ gives a homotpoy equivalence of fibers, and so at least $\pi_n(F) \simeq \pi_{n}(F')$, but I can't quite show that they both give the same class in cohomology.

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Well, any homotopy equivalence maps a fundamental class to a fundamental class; and the rest follows immediately from functoriality of spectral sequence, no? –  Grigory M Sep 21 '11 at 5:02
    
@GrigoryM - It would seem like that would work! I guess I just need to prove that first statement then –  Juan S Sep 21 '11 at 6:19

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