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Let $L: \mathcal C^\to_\leftarrow L\mathcal C : i$ be an adjunction with $i$ fully faithful. In ordinary category theory, $L$ is left exact iff the class of $L$-local morphisms is stable under base change [1].

This appears to be true $\infty$-categorically, as well. Is there a proof in the literature?

But $\infty$-categorically, this is no longer true:

Example: Let $\mathcal C$ be the $\infty$-category of spaces and $L\mathcal C$ the full subcategory of $n$-truncated spaces for some fixed $n$, so that $L$ is the $n$-truncation functor and the $L$-local morphisms are those with $(n+1)$-connected fibers. Then $L$ is not left exact (failing to preserve, for example, the pullback square $K(\mathbb Z, n) \rightrightarrows \ast, \ast \rightrightarrows K(\mathbb Z, n+1)$), but the $L$-local morphisms are stable under base change.

Question: Let $L: \mathcal C^\to_\leftarrow L\mathcal C: i$ be an adjunction of finitely-complete $\infty$-categories with $i$ fully faithful. Let $\mathcal W = L^{-1}(\{\textrm{isos}\}) \subseteq \textrm{Mor} \mathcal C$ be the class of $L$-local morphisms. What are necessary and sufficient closure conditions on $\mathcal W$ ensuring that $L$ is left exact?

I'm happy to assume that $\mathcal C$ is presentable, or even an $\infty$-topos, and that $\mathcal W$ is of small generation.

[1] Here we assume that $\mathcal C$ is finitely complete. A morphism $f$ is said to be $L$-local if $L(f)$ is an isomorphism, and a class of morphisms $\mathcal W$ is stable under base change if $f \in \mathcal W$ implies $f' \in \mathcal W$ where $f'$ is any pullback of $f$ along an arbitrary morphism.

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  • $\begingroup$ Is the class of morphisms of spaces inverted by $n$-truncation stable under pullback? Isn't your pullback square a counterexample?: $\ast \to K(\mathbb{Z},n+1)$ is inverted by $n$-truncation, but $K(\mathbb{Z},n) \to \ast$ isn't, right? $\endgroup$ – Alexander Campbell Jan 8 at 21:21
  • $\begingroup$ @AlexanderCampbell -- yes, I was getting confused. This criterion does hold $\infty$-categorically. I've just written up a proof over here. $\endgroup$ – Tim Campion Jan 8 at 21:30
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Unless I misunderstand the statement, this is precisely proposition 6.2.1.1 in Higher Topos Theory.

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