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Mar 13, 2019 at 13:09 vote accept Mike Shulman
Mar 13, 2019 at 13:03 comment added Mike Shulman @Denis-CharlesCisinski As far as I know that is only true if you either remove the $\lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $\lambda\lhd\mu$ (which changes it from "for sufficiently large $\mu$" to "for arbitrarily large $\mu$" -- see Remark 2.15 in AR).
Mar 13, 2019 at 8:47 history became hot network question
Mar 13, 2019 at 7:50 answer added Jiří Rosický timeline score: 13
Mar 13, 2019 at 7:26 comment added D.-C. Cisinski We may assume that $F$ preserves small $\lambda$-filtered colimits. Isn’t it true that, for $\mu$ large enough, an object is $\mu$-presentable if and only if it is a $\mu$-small $\lambda$-filtered colimit of $\lambda$-presentable objects? Another way to put it, is that for $\mu$ large enough (e.g. larger than $\lambda$ and than the set of maps between any two $\lambda$-presentable objects), the property of $\mu$-presentability of an object $X$ is simply the fact that the set of maps from a $\lambda$-presentable object to $X$ is of cardinal $\leq\mu$.
Mar 13, 2019 at 5:17 comment added Reid Barton For fixed $\alpha$, does $\mu^\alpha = \mu$ hold for all sufficiently large regular $\mu$?
Mar 13, 2019 at 2:25 history asked Mike Shulman CC BY-SA 4.0