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Fix a universe $\mathcal{U}$. Call a category $\mathcal{U}$-complete if every diagram indexed by a $\mathcal{U}$-small category has a limit, and a functor $\mathcal{U}$-continuous if it preserves $\mathcal{U}$-small limits. Usually, when one fixes a universe, one calls this simply complete and continuous.

Now assume we are given $\mathcal{U}$-complete categories $C,D$ and a $\mathcal{U}$-continuous functor $F : C \to D$. Does then $F$ also preserve limits (which exist in $C$) which are not necessarily $\mathcal{U}$-small?

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up vote 4 down vote accepted

No. I'll answer for the case of colimits as follows: Consider for example the category $O$ of ordinals (as a poset), and adjoin a terminal object $T$, making a larger category $C$. Then this terminal object is a (large) colimit over the diagram $O\to C$. However, a cocontinuous functor $C\to D$ can send $T$ anywhere that admits a cone under the diagram $O\to D$, not necessarily its colimit (if the colimit even exists).

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Right! Thank you – Martin Brandenburg Feb 17 '11 at 16:07

The answer is yes if you put in some additional smallness/accessibility/presentability assumptions. For example, if $C$ is the $U$-cocompletion of a $U$-small subcategory, each of whose objects are ($U$-small)-presentable, then any $U$-cocontinuous functor is a left adjoint, and hence absolutely cocontinuous.

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Interesting! Actually this situation seems to arise very often ... – Martin Brandenburg Feb 17 '11 at 18:40
@Martin: Yes, I think it's the standard situation, and Owen's answer is the standard counterexample to the more general claim. In other news, I'm only recently a big fan of presentable/accessible categories, but I have become a big fan of them, so I hope you don't mind that I answer most of your questions with reference to them. – Theo Johnson-Freyd Feb 17 '11 at 23:15
Au contraire, I like your answers. – Martin Brandenburg Feb 18 '11 at 8:56

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