While reading the Hovey Model Category I learned that $I^{rlr} = I^r$ and $I^{lrl} = I^l$ for every class of morphisms $I$. What other relations are there between the operations of taking (weak) orthogonals? Are all $I^l, I^{ll}, I^{lll}, I^{llll}, ..$ distinct in general?
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asked Sep 6, 2022 at 3:53
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2$\begingroup$ In general no: in any category $C$, for $I$ the class of its isomorphisms, $I^l=\hom(C)$ (all morphisms of $C$), and $\hom(C)^l=I$, so you have an "oscillating" sequence. $\endgroup$– foscoCommented Sep 6, 2022 at 8:48
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3$\begingroup$ Another potential source of counterexample is $Set$: in the category of sets (with choice and all you need to make this true) $Epi=Mono^{l!}$ (strong orthogonality) but also $Epi=Mono^r$ (because $(Epi,Mono)$ is an orthogonal FS on $Set$, but $(Mono,Epi)$ is a weak one), so $Epi^l=Mono^{rl}=Mono$; now $Epi^{ll}=Mono^l$ which I would try to show is just $Epi=Mono^{l!}$, despite the stronger orthogonality request. $\endgroup$– foscoCommented Sep 6, 2022 at 8:55
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$\begingroup$ Hmm, I use "in general" in the opposite sense. I mean "are there any relations true for every class I", "what are the equalities we are sure of, regardless of information about the class of morphisms". $\endgroup$– Arshak AivazianCommented Sep 7, 2022 at 8:50
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$\begingroup$ @fosco Note that if $M$ is a class of monomorphisms in a 1-category, then $M^{l} = M^{l!}$ always, so it's not necessary to distinguish between weak and strong orthogonality in this particular case. So we do indeed have $Epi = Mone^l$ in $Set$, and get another oscillating sequence as you suggested! $\endgroup$– Tim CampionCommented Sep 9, 2022 at 11:48
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$\begingroup$ I would suggest that typically it's not very natural to consider the operations $(-)^{ll}$ or $(-)^{rr}$. Let $\mathcal L = \mathcal R = \{\text{classes of morphsisms in }\mathcal C\}$, ordered by inclusion. Then $(-)^l : \mathcal R^{op} {}^\to_\leftarrow : \mathcal R (-)^r$ is an adjunction, with interesting fixed points. But I think of it as "an accident" that $\mathcal L$ and $\mathcal R$ are the same set. That being said, I wouldn't rule out that this perspective might be misleading -- maybe there are interesting things to say here! $\endgroup$– Tim CampionCommented Sep 9, 2022 at 11:54
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