Is there a functor that preserves all small limits but not a large one?
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$\begingroup$ I did not know that some limits are small and some are large. $\endgroup$– Wadim ZudilinCommented Jul 26, 2010 at 12:48
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1$\begingroup$ "small" refers to the index category of the diagram; this should be small (in the sense that its object class is a set). $\endgroup$– Martin BrandenburgCommented Jul 26, 2010 at 13:00
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1$\begingroup$ By the way, I've never seen so far a large limit which exists and does not restrict somehow to a small limit. $\endgroup$– Martin BrandenburgCommented Jul 26, 2010 at 13:02
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1 Answer
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Let's try this. I'll use colimits, so take the opposite.
The class of all ordinals is ordered. Add one more element $\infty$ at the end, bigger than all of them. View this "large ordered set" as a large category $\cal C$. A small diagram in $\cal C$ has colimit $\infty$ if $\infty$ occurs in the diagram, and otherwise it has a colimit less than $\infty$. But the large diagram consisting of everything except $\infty$ has colimit $\infty$. The functor to the ordered set $\lbrace 0<1\rbrace$ that sends $\infty$ to $1$ and everything else to $0$ preserves small colimits but not all colimits.
answered Jul 26, 2010 at 16:26
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$\begingroup$ 1+. Nice example. Somehow it is 'generic'. $\endgroup$ Commented Jul 26, 2010 at 17:03
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$\begingroup$ Yes. First I tried to come up with a clever counterexample, but then I found this instructive one instead. $\endgroup$ Commented Jul 27, 2010 at 0:57
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$\begingroup$ Sweet, didn't know there's such a simple example. To Tom Goodwillie and Martin Brandenburg, why do so many people use proper class? I follow MacLane's method and think of small being an element of a fixed universe U, and large as a set that's not in U. They are all "sets" in this way, and the class of all ordinals can be interpreted as the set of all small ordinals. If you insist on using proper class, how can you construct functor categories like Set^Set? My set theory is very weak but I'm pretty sure talking about Set^Set is like talking about the power class of the class of all sets. $\endgroup$– AmadeusCommented Jul 27, 2010 at 4:32