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.

  • $\begingroup$ 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? $\endgroup$ Commented Jan 8, 2020 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
    Commented Jan 8, 2020 at 21:30

1 Answer 1


Unless I misunderstand the statement, this is precisely proposition in Higher Topos Theory.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.