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It seems to be the case that filtered colimits commute with finite limits in the category Set (for instance, this is shown in Why do filtered colimits commute with finite limits?), but does the same hold for the category of pointed sets?

For the case of when the finite limit is a binary product, the OP in the linked post explains why the claim holds in Set and generally any Cartesian closed category with filtered colimits. But the category of pointed sets is not Cartesian closed (namely because the category of pointed sets has a zero object and the only Cartesian closed category with a zero object is the trivial one).

So, first, is it even true that filtered colimits commute with binary products in the category of pointed sets (if not, is there an easy counterexample?), and if so, can we replace "binary products" with any "finite limit"?

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Yes, filtered colimits commute with finite limits in the category of pointed sets. This is because the forgetful functor from the category of pointed sets to the category of sets creates finite limits and filtered colimits -- in fact, it creates all limits and all connected colimits -- and so the category of pointed sets inherits this property from the category of sets.

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I find it sometimes useful to remember that the stated commutation property holds in any ind-category. Pointed sets is ind-(finite pointed sets). Of course, to verify your particular case by hand, working with elements, literally takes 5 minutes. But who wants to waste five minutes on elements? 😛

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    $\begingroup$ And so, in particular, it holds in the category of models of any finitary algebraic (or essentially algebraic) theory, that is, one described by operations that take only finitely many arguments. $\endgroup$ – Reid Barton Oct 26 at 14:16

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