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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.

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(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.

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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.)

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  • $\begingroup$ Do you mean to have $\Omega B$ on the top right corner of your second square? $\endgroup$ – Denis Nardin 2 days ago
  • $\begingroup$ Whoops, yes, thanks for spotting the typo @DenisNardin. $\endgroup$ – skd 2 days ago

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