Assume that we have a pullback square $$ \begin{array}{ccc} A & \rightarrow & B \\ \downarrow & & \downarrow \\ C & \rightarrow & D \\ \end{array} $$ with all functors accessible, and all categories presentable (if needed, assume that $A \to C$ and $B \to D$ are left adjoints).

Suppose further that we have three saturated (that is, closed under retracts, pushouts and transfinite compositions) classes of arrows $S(B),S(C),S(D)$ of, respectively, $B,C,D$. Assume that the functors $B \to D, C \to D$ preserve these classes. (It seems likely to me that the set $S(D)$ should not play any importance in what follows)

Form the set $S(A)$ by requiring that it consists of all arrows of $A$ that are sent to $S(B)$ and $S(C)$ by the respective functors.

Question: if $S(B),S(C),S(D)$ are generated by (small) sets, does it imply that $S(A)$ is generated by a set as well?

EDIT: it may be not obvious that the formed class S(A) is weakly saturated. Assume it is.

EDIT2: I do not know if the original question admits an affirmative answer. What is true is that, following a remark of Tim to this post, one can consider the situation with the right classes. Let me leave a statement that may be of use for someone who googles this post up.

Lemma. Let $A$ be a presentable category together with a weak factorisation system $(L,R)$ having the property that

  1. $(L,R)$ is functorial (a closer inspection may be used to drop this),
  2. The functor $A^{[1]} \to A^{[2]}$ associated to the weak factorisation system is accessible,
  3. The right class $R$ viewed as a subcategory of $A^{[1]}$ is accessible and accessibly embedded.

Then there exists a set $L_0 \subset L$ generating $(L,R)$.

  • $\begingroup$ No colimit preservation for $C\to D$ might be assumed? $\endgroup$ Jan 27, 2018 at 13:21
  • $\begingroup$ Ideally, no. The example I have in mind is the category of sections of a Grothendieck fibration $E \to [1]$ such that $E(0),E(1)$ are presentable and $E(1) \to E(0)$ is accessible, with no evident adjoint. $\endgroup$
    – Edouard
    Jan 27, 2018 at 19:09
  • 2
    $\begingroup$ The hypotheses feel a bit awkward, suggesting to me that this is not the right abstraction to work with. However, if you look at the corresponding right classes $R(B),R(C)$ of of these wfs, then if you define $R(A) = R(B) \times_{D^{[1]}} R(C)$ it will be accessible and accessibly embedded (as the pullback of such along accessible functors). So assuming that $R(A)$ is again the right class of a wfs, the left class corresponding to $R(A)$ will be small-generated. $\endgroup$
    – Tim Campion
    Jan 27, 2018 at 21:44
  • $\begingroup$ Tim: thank you, I think your comment solves one particular case I am looking in. $\endgroup$
    – Edouard
    Jan 28, 2018 at 12:14
  • $\begingroup$ @TimCampion could you please provide a reference for the statement that you mention? Namely that an accessible and accessibly embedded subcategory of $A^{[1]}$ is generated by a set of morphisms of $A$. I only find one for small injectivity-classes, which does not quite give the same. Thank you. $\endgroup$
    – Edouard
    Jan 30, 2018 at 20:15

1 Answer 1


Yes, see Theorem 3.2 in Makkai and Rosický's Cellular categories.

  • $\begingroup$ Thanks for the reference. They seem to assume that both $C \to D$ and $B \to D$ preserve colimits, which is not exactly my case. Do you reckon that the proof goes through still? $\endgroup$
    – Edouard
    Jan 28, 2018 at 12:04
  • 2
    $\begingroup$ I don't think S(A) is always weakly saturated under such conditions. If you do know it is weakly saturated, then Tim Campion's argument should suffice. $\endgroup$ Jan 28, 2018 at 19:56

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