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Suppose $D$ is an $\infty$-category, then we have the equivalence $$ \text{Fib} (D) \substack{ \text{St} \\ \longrightarrow \\ \cong \\ \longleftarrow \\ \text{Un}} [ D^\text{op}, \mathbf{Kan}]$$ between (right) fibrations over $D$ and functors from the opposite of $D$ to Kan complexes ($\infty$-groupoids) via the Grothendieck construction.

Given a map $f: C \to D$ of $\infty$-categories, there is a functor $f^* : \text{Fib} (D) \to \text{Fib}(C)$ by forming pullbacks. Since $f$ induces a functor $C^\text{opp} \to D^\text{opp}$, there is also a functor $f^* : [ D^\text{op}, \mathbf{Kan}] \to [ C^\text{op}, \mathbf{Kan}]$.

Question: For a fibration $\pi : X \to D$ do we have $$ \text{St} f^* \pi \cong f^* \text{St} \pi \quad ?$$ So is the Grothendieck construction compatible with pullbacks?

Thanks for any hints.

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Yes, though it is usually written as the commutativity of unstraightening with pullback (on the $\infty$-categorical level it doesn't matter, since straightening and unstraightening are inverse equivalences). This compatibility even holds on the point-set level if one uses a suitable model categorical presentation of the straightening-unstraightening equivalence, see Proposition 2.2.1.1 of higher topos theory (if you combine (1) and (2) of that proposition then you get the compatibility of straightening with left Kan extensions. The compatibility of unstraightening and pullbacks is obtained by switching to right adjoints).

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  • $\begingroup$ I tried to make this precise, but this does not really give me the compatibility of pullback and straightening up to isomorphism, right? Can you maybe elaborate? Thanks in advance. $\endgroup$
    – Lukas Woike
    Commented Jun 20, 2018 at 20:18
  • $\begingroup$ Hi Lukas, I was wondering the same and came up with a plausible argument. Let's see if I understand correctly - otherwise, maybe Yonatan can elucidate the matter. HTT 2.2.1.1.(1)+(2) in the $\infty$-categorical setting says that $\mathrm{St} \circ f_{!} \simeq \mathrm{LKan}_f \circ \mathrm{St}$. Taking adjoints of both sides yields $f^\ast \circ \mathrm{Un} \simeq \mathrm{Un} \circ f^\ast$, where the first $f^\ast$ is pullback along $f$ and the second is precomposition by $f$. Now conjugate by $\mathrm{St}$ to get $\mathrm{St} \circ f^\ast \simeq f^\ast \circ \mathrm{St}$. $\endgroup$
    – Lorenzo Riva
    Commented Feb 24, 2023 at 5:24

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