Theorem 1.17 of Emily Riehl's Factorization Systems says that given a class of maps $\mathcal M$, $\mathcal M^\perp$ is closed under limits and dually $^\perp\mathcal M$ is closed under colimits.

The argument for the former is as follows.

Given two arrows $e,m$, the square below always commutes.

$$\require{AMScd} \begin{CD} \mathsf{Hom}(B,X) @>{e^\ast}>> \mathsf{Hom}(A,X)\\ @V{m_\ast}VV @VV{m_\ast}V\\ \mathsf{Hom}(B,Y) @>>{e^\ast}> \mathsf{Hom}(A,Y) \end{CD}$$

This assignment is functorial in $e,m$, yielding a functor

$$S:(\mathsf C^\text{op})^2\times \mathsf C^2\longrightarrow \mathsf{Set}^{2\times 2}$$

that is continuous in each variable. $m\in \mathcal M^\perp$ iff $S(e,m)$ is a pullback for all $e\in M$. The full subcategory of the codomain spanned by pullback squares is closed under limits, completing the proof.

Why is $S$ continuous in each variable? Why is the stated full subcategory closed under limits?

I'm guessing the former has to do with Yoneda and the latter is simply commutation of limits with limits, but I'd like to make sure.

Added. Having already received an answer (which conflicts with the stated theorem), I asked to have this question migrated to MO in order to receive more input. I'm hesitant to accept the author has stated a false theorem, but do not see why $S$ is continuous in each variable.

Added. Am I correct in saying moreover that Theorem 1.17 holds in prefactorization systems in general?

  • $\begingroup$ sure, prefactorization systems are fine $\endgroup$ – Tim Campion Aug 11 '16 at 3:27
  • $\begingroup$ @TimCampion do you by any chance understand the proof? $\endgroup$ – Arrow Aug 11 '16 at 6:41
  • $\begingroup$ I think it works with user54748's fix. In any event, you can prove it directly by a diagram chase. Orthogonality is all that's required, not factorizations. $\endgroup$ – Tim Campion Aug 11 '16 at 20:46

First of all, unless I'm overlooking something, $S$ preserves only pointwise limits in each argument, I'll write down a counterexample at the end of the post. To show that it preserves those fix for example a morphism $e : A → B$ in $C$ to be the first argument of $S$, and denote the resulting functor $C^I → \mathrm{Set}^{I×I}$ with $S'$. (I'm using $I$ for the interval category here instead of $2$ to avoid ambiguity.) If $m : X → Y$ is the pointwise limit of a diagram of $m_i : X_i → Y_i$'s you need to prove that $S'(m)$ is the limit of $S'(m_i)$'s. But limits in $\mathrm{Set}^{I × I}$ are calculated pointwise, so you only need to show that every vertex of $S'(m)$ is the limit of the corresponding vertices of $S'(m_i)$ and ditto for the edges, and these all are just by the continuity of the covariant hom.

As for the second part, yes, it's just commutation of limits with limits. A limit of squares in the codomain is calculated pointwise, and if every square in a limit diagram is pullback, so is the limiting square.

To see that $S$ doesn't necessarily preserve limits in $C^I$ that aren't pointwise, take $C$ to be the subposet $\{\{1\}, \{5\}, \{1,2\}, \{1,3\}, \{1,4\}, \{1,2,3,5\}, \{1,3,4,5\}\}$ of $(\mathcal P(\mathbb N), ⊆)$. If you draw this poset, you will see that $h : \{1\} ⊆ \{1, 3\}$ is the product of $f : \{1,2\} ⊆ \{1,2,3,5\}$ and $g : \{1,4\} ⊆ \{1,3,4,5\}$ in $C^I$ (it's the only cone over them), but it isn't a pointwise product (since $\{1,2,3,5\}$ and $\{1,3,4,5\}$ don't have a meet in $C$). (This example is exercise 2.17.9 in Borceux's Handbook). Furthermore, $H = \mathrm{Hom}(\{5\}, -)^I$ doesn't preserve this limit, so $S(\mathrm{id}, -)$ can't presrve it either: $Hh$ is $∅ → ∅$, while $Hf$ and $Hg$ are isomorphic to $∅ → \{*\}$.

This doesn't seem to be much of a problem, however. If $C$ is complete, as I initially thought is assumed, then all limits in $C^I$ are pointwise, and if they aren't, I'm not sure of how much interest those that aren't pointwise would be.

  • $\begingroup$ Could you add some more details about the factorization of $S$ in each variable? $\endgroup$ – Arrow Jul 28 '16 at 20:23
  • $\begingroup$ @Arrow: Actually I couldn't, what I wrote is sloppy and wrong, my apologies for that. I'll replace this with something more appropriate tomorrow. $\endgroup$ – user54748 Jul 29 '16 at 2:53
  • $\begingroup$ The statement of the theorem involves no completeness hypotheses on the ambient category, so I'm a bit hesitant to accept your answer.. I'll ask to migrate this to MO and see what response this questions finds there. $\endgroup$ – Arrow Jul 30 '16 at 8:34
  • $\begingroup$ @Arrow: No problem, that's probably the best course of action. $\endgroup$ – user54748 Jul 30 '16 at 12:31
  • $\begingroup$ I think there's a typo: the counterexample is Exercise 2.17.9 in volume I of my edition Borceux's handbook. $\endgroup$ – Tim Campion Aug 11 '16 at 11:58

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