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Oct 30, 2015 at 8:49 vote accept Daniel Grady
Oct 28, 2015 at 11:51 comment added Zhen Lin Of course, as in SGA, we may replace the assumption that $f : C \to D$ preserves finite limits with the assumption that $f_! : \mathrm P (C) \to \mathrm P (D)$ does (or, better, that $a_D f_! : \mathrm P (C) \to \mathrm{Sh} (D)$ does) when $C$ does not have all finite limits.
Oct 28, 2015 at 11:05 comment added Daniel Grady ah, perfect. I thought the argument might go the same way as in the classical case but I wasn't completely sure. Thanks for the response.
Oct 28, 2015 at 10:52 history edited AAK CC BY-SA 3.0
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Oct 28, 2015 at 10:47 history answered AAK CC BY-SA 3.0