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Let $\mathcal{A} \subset \mathcal{K}$ be two locally presentable categories. $\mathcal{A}$ reflective and closed under filtered colimits. Then $\mathcal{A}$ is a small-orthogonality class. Let $R:\mathcal{K}\to \mathcal{A}$ be the reflection. Let $G$ be a dense generator of $\mathcal{K}$ consisting of objects not belonging to $\mathcal{A}$. Can we conclude that $\mathcal{A}$ is the small-orthogonality class with respect to the set of maps $\{\eta_g:g\to Rg \mid g\in G\}$ ?

I think that the answer is negative in full generality and I would like to see a counterexample.

Note: I have edited my question to remove the case $G\subset \mathcal{A}$.

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    $\begingroup$ Just to clarify, by orthogonal, do you mean strongly orthogonal (unique lifts) or weakly orthogonal (just existence of lifts)? I guess the former, but a few authors write orthogonal for the latter. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jul 15, 2017 at 10:23
  • $\begingroup$ In my answer is crucial to have strong orthogonality. $\endgroup$
    – Ivan Di Liberti
    Commented Jul 15, 2017 at 10:41
  • $\begingroup$ @PeterLeFanuLumsdaine I mean strong orthogonality, otherwise I use the word injectivity. $\endgroup$
    – Philippe Gaucher
    Commented Jul 15, 2017 at 15:39

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It seems to me that a negative answer follows immediately from the possibility, in many cases, of choosing $G\subset \mathcal A$, so that $G$ doesn't know anything about $R$ until one closes it under some colimits. Given such a $G$, every object of $\mathcal K$ is right orthogonal to each $\eta_g$, since these are isomorphisms.

For instance, let $\mathcal A$ be torsion-free abelian groups, which are reflective and closed under filtered colimits in abelian groups, $\mathcal K$. Now $\mathcal K$ has a dense generator consisting of torsion-free groups; for instance, the singleton $\mathbb{Z}^2$ suffices.

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  • $\begingroup$ I perfectly understand your point. The motivation of the question is that in the particular situation I am working on, it is true. And I cannot figure out what $G$ has special. To prove it, I have to make boring annoying (but easy) calculations. I was just wondering whether I was not missing a categorical argument. In my situation, the canonical diagram with respect to $G$ is not even filtered. Maybe there is no way for me to avoid to make the calculations. $\endgroup$
    – Philippe Gaucher
    Commented Jul 14, 2017 at 19:34
  • $\begingroup$ @PhilippeGaucher Well, at least in the additive case I was discussing, it seems relevant that $\mathcal A$ sits in a recollement in which $G$ is sent to zero on the other side. Perhaps this would work if $G$ is sent to a(dense?) generator on both sides of a recollement? I'm not sure what the non-additive version of this would be, if anything. $\endgroup$
    – Kevin Carlson
    Commented Jul 15, 2017 at 1:11
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    $\begingroup$ Theorem 1.39 in my book with Adámek is not correct for $\lambda=\omega$ (see my paper with Adámek and Hébert More on orthogonality in locally presentable categories, Cah. Top. Géom. Diff. Cat. XLII (2001), 51-80). Counterexamples there also serve for the question of Philippe. But the theorem is valid for $\lambda$ uncountable (see my paper with Hébert Uncountable orthogonality is a closure property, Bull. London Math. Soc. 33 (2001), 685-688). $\endgroup$
    – Jiří Rosický
    Commented Jul 16, 2017 at 8:15
  • $\begingroup$ @JiříRosický It may be useful to point out that an erratum for your book is available here: tu-braunschweig.de/Medien-DB/iti/cor.pdf $\endgroup$
    – Philippe Gaucher
    Commented Jul 19, 2017 at 8:59

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