Action of fundamental group on homotopy fiber

For a Serre fibration of pointed topological spaces $$f:X \to B$$, there is an action of $$\pi_1\left(B,b_0\right)$$ on the fiber $$F$$. The construction of this action I'm familiar with uses a lift $$F\times I \to X$$ of the map $$F \times I \xrightarrow{\pi} I \xrightarrow{\gamma} B$$ for any $$\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$$.

Now for a general map between two $$\infty$$-groupoids $$f:X \to B$$, we can use some version of the Grothendieck construction to construct a map $$\phi_f : B \to \operatorname{Grp}_\infty$$, and then for an element $$\left[\gamma \right] \in \pi_1 \left(B,b_0 \right)$$, $$\phi_f \left(\gamma \right)$$ is an automorphism of $$\phi_f \left(b_0 \right)$$, which, I guess, generalizes the previous definition (please correct me if that is already false). Is there an explicit description of this automorphism for a specific $$\gamma$$? (i.e. in terms of pullbacks / pushouts / sections of the maps $$f,\gamma$$ etc.) I'm particularly interested in writing down obstructions for triviality of this action.

(This answer is written in a model-independent fashion -- translate to your favourite formalism).

For every path $$\gamma:[0,1]\to B$$ you get an isomorphism in the homotopy category $$X_{\gamma0}\xrightarrow{\sim} X_{\gamma1}$$ (where with $$X_b$$ I denote the homotopy fiber over $$b\in B$$). Probably the easiest and most geometric way of constructing it is to consider the space of lifts.

Let $$\operatorname{Sec}_\gamma(f)$$ be the space whose objects are sections up to homotopy over $$\gamma$$ $$\operatorname{Sec}_\gamma(f)=\{(\tilde\gamma,H)\mid \tilde\gamma:[0,1]\to X,\ H:f\tilde\gamma\sim \gamma\}$$ (this is nothing else but the homotopy fiber over $$\gamma$$ of the map $$X^{[0,1]}\to B^{[0,1]}$$). Then you have a zig zag of maps $$X_{\gamma0}\xleftarrow{ev_0} \operatorname{Sec}_\gamma(f)\xrightarrow{ev_1} X_{\gamma1}$$ where the two maps are evaluation at 0 and 1 respectively. Both maps are homotopy equivalences (this requires some proof, but it's not terribily hard: they're both trivial fibrations when $$f$$ is a fibration), and so you can define the map in the homotopy category as $$ev_1 \circ ev_0^{-1}$$.

In order to prove that the action of a loop is trivial, you'll have to prove that $$ev_0$$ and $$ev_1$$ are homotopic. I'm not aware of a general way of attacking this problem, but of course studying the behaviour of the two maps on various algebraic invariants can often provide obstructions.

Let $$f:E\to B$$ be a map of based spaces, and let $$F$$ be the homotopy fiber. Here is another way of constructing the action of $$\Omega B$$ on $$F$$. By definition, there is a homotopy pullback square $$\require{AMScd} \begin{CD} F @>>> \ast\\ @VVV @VVV \\ E @>>> B.\\ \end{CD}$$ Taking homotopy pullbacks along the inclusion $$\ast\to B$$ produces a map to the above homotopy pullback square from the following one: $$\require{AMScd} \begin{CD} \Omega B\times F @>>> \Omega B\\ @VVV @VVV \\ F @>>> \ast.\\ \end{CD}$$ The two morphisms in this square are the projections. The action of $$\Omega B$$ on $$F$$ is just the map between the top left corners of these squares; let's call this map $$\mu$$. This action is not just the projection: this construction shows that there is a homotopy pullback square $$\require{AMScd} \begin{CD} \Omega B\times F @>{\mathrm{pr}}>> F\\ @V{\mu}VV @VVV \\ F @>>> E;\\ \end{CD}$$ if $$\mu$$ was just projection onto $$F$$, then the space $$E$$ in the bottom right corner would have to be replaced by $$F\times B$$. (Note that this diagram shows that the composite $$\Omega B \times F\to F\to E$$ is trivial on $$\Omega B$$. You can also see this by the explicit model of this map in spaces: this composite just sends a pair $$(\gamma, [e\in E, p:\ast\to f(e)])$$ to $$e$$.) I don't know of general methods to show that the action is trivial. (Remark: one potential advantage of phrasing the construction in this way is that it works in any ($$\infty$$-)category with finite homotopy limits.)

• Do you mean to have $\Omega B$ on the top right corner of your second square? May 23, 2020 at 20:29
• Whoops, yes, thanks for spotting the typo @DenisNardin.
– skd
May 23, 2020 at 21:10