# the group of fibre homotopy classes of fibre homotopy equivalences. how to compute it

As the title suggest, for a fibration $\xi \colon E \to B$ I need to compute the group $$Aut(\xi)$$ which has as elements the fibre homotopy classes of fibre homotopy equivalences $E\to E$ for this trivial fibration: $$K(\pi,1) \times BSO \xrightarrow{\pi_2} BSO$$ where $K(\pi,1)$ is the Eilenberg-MacLane space with $\pi_1=\pi$ and $BSO$ is the classifying space for $SO$. This is the suggested computation: $$[K(\pi,1) \times BSO,K(\pi,1)] \simeq [K(\pi,1),K(\pi,1)] \simeq Out(\pi)$$ where $[K(\pi,1) \times BSO,K(\pi,1)]$ is justified by the fact that we are requiring that the map preserve the fibration and therefore the second component has to be trivial (i.e. this is the only interesting component). The next step is justified by the fact that we know up to homotopy that maps into a $K(\pi,1)$ form a connected space are in bijection with maps $\pi_1(K(\pi,1)\times BSO)\to \pi$, together with $\pi_1BSO=*$. Last passage is another application of this reasoning together with the fact that inner automorphisms correspond to base point changes and an element of $[K(\pi,1),K(\pi,1)]$ is independent of base point.

Everything seems to work a part from the fact that according to the definition we are looking for fibre homotopy classes of (fibre) homotopy equivalences. I put the bracket around the second fibre since it's known that we can relax such hypothesis if we are dealing with fibre homotopy equivalences.

According to me we are computing homotopy classes of such fibre homotopy equivalences, and not the "fibre" ones. Am I missing something?

• The very first justification is where you drop the need for these maps to be fiber-preserving. From a map $K(\pi,1) \times BSO \to K(\pi,1)$, you may produce a fiber-preserving map $K(\pi,1) \times BSO \to K(\pi,1) \times BSO$ and vice versa. – user98602 Sep 20 '16 at 17:42
• Dear Mike, does this trick work even for the homotopies? Because the maps are clearly fibre preserving, my problems lies with the homotopies between them. It is possible that I misunderstood your comment – Luigi M Sep 20 '16 at 17:59
• Yes. I'm not sure what there is to say... fiber preserving homotopies of the latter correspond bijectively to homotopies of the former. – user98602 Sep 20 '16 at 18:01
• Yes I realized what you meant just now. If you make an answer with your comment I'd be happy to close this question – Luigi M Sep 20 '16 at 18:03

Given a trivial bundle $B \times F \to B$, fiber-preserving maps $B \times F \to B \times F$ that cover the identity are in canonical bijection with maps $B \times F \to F$, by $f \mapsto \pi_2 f$, and in the other direction, $g \mapsto \pi_1 \times g$. Setting $F = K(\pi,1)$ and $B = BSO$ (or $BSO \times I$ in the case of homotopies) we see that there is a canonical bijection between fiber-preserving maps up to fiber-preserving homotopy of $F \times B \to F \times B$ and elements of $[F \times B, F]$.