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Feb 10, 2020 at 15:59 comment added Tim Campion Haha! I can't believe I forgot such a striking fact!
Feb 10, 2020 at 14:56 comment added Kevin Carlson Nope! As, er, tcamps recalls in the comments here: math.stackexchange.com/a/2128004/31228... the functor is roughly “the set of split subobjects with a disjoint basepoint.” In particular any object lacking nontrivial split subobjects will be mapped to a two-point set, so the functor is not injective on isomorphism classes even when applied to Set itself. If there are a large number of such objects, then the functor will have a large fiber on isomorphism classes while still being conservative.
Feb 10, 2020 at 7:54 comment added Tim Campion @KevinCarlson Whoa -- that's surprising! When you say "every locally small $C$ admits a functor $C \to Set$", do we have to assume that $C$ has no more isomorphism classes of objects than the size of the universe?
Feb 10, 2020 at 7:32 comment added Kevin Carlson Fun fact, in case you intended the implication in your last point that $h\mathcal S$ admits no conservative functor whatsoever to Set: actually every locally small category admits such a functor! I didn’t believe that when I heard it, but it’s yet another coup de Freyd. Of course there is still a big gap between $h\mathcal S$ and $D(R)$ in admitting such a functor which is a coproduct of representables.
Feb 10, 2020 at 4:52 history edited Tim Campion CC BY-SA 4.0
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Feb 10, 2020 at 4:21 history edited Tim Campion CC BY-SA 4.0
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Feb 10, 2020 at 4:12 history edited Tim Campion CC BY-SA 4.0
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Feb 10, 2020 at 4:02 history edited Tim Campion CC BY-SA 4.0
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Feb 10, 2020 at 3:48 history edited Tim Campion CC BY-SA 4.0
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Feb 10, 2020 at 3:42 history answered Tim Campion CC BY-SA 4.0