Tim Porter wrote:
Dear All, There is an interesting old result on fibrations that may shed light on this. It is well known that if $F\to E\to B$ is a fibration of pointed spaces then $\pi_1 F \to \pi_1 E$ is a crossed module and in fact any crossed module can be given in this form. Now one tries to prove it. First the action: take a loop g in E and another a in F. concatentate to get gag^{-1}. This is a loop in E whose image in the base,B, is null homotopic. Pick a null homotopy that does this. Lift it to a homotopy in E starting at gag^{-1} using the homotopy lifting property of the fibration. Evaluate the other end of the lift. This is a loop in F. The corresponding element of \pi_1 F is the result of acting on the class of a by the class of g. Note the way the action is determined up to homotopy. The verification that the rules work up to homotopy is left as an exercise.
I learnt this from a paper by Eric Friedlander, who attributed it to Deligne. I suspect it is already essentially in Whitehead's Combinatorial Homotopy II paper or Peter Hilton's lovely little book on Homotopy Theory.
It suggests a `homotopy everything' version of crossed module, not just a lax one.
There is a closely related homotopy everything version which I bet can be adapted to this problem. Consider \Omega B, the based loop space and F as the fibre over that base point. Then an argument like that above gives and action of \Omega B on F which satisfies the usual rules for an actin only up to homotopy Easiest way to say it is the adjoint map \MOmega B --> Aut F meaning the self homotopy equivalences of F is a strongly homotopy associative map of monids (if you use Moore loops) or of A_\infty spaces. It should be in "Parallel transport in fibre spaces," Bol. Soc. Mat. Mexicana (1968), 68-86. or "Associated fibre spaces," Michigan Math. Journal 15 (1968), 457-470. Hope that helps jim
Its advantage is that it clearly links up the structure with the quite classical topological version of fibrations and so should be adaptable to other situations.
Hope this helps.
Tim