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Nathan Corbyn (Feb 22 2024 at 17:38):If F : C -> D preserves finite products and equalisers that exist in C, does F necessarily preserve all finite limits that exist in C? (I know this is true if C is finitely complete, but what about the general case? My assumption is no, but I'm struggling to think of a counter-example.)
Ralph Sarkis (Feb 22 2024 at 17:57):Why are you assuming it is not true? What do you think works in the case where is finitely complete and does not work when it is not?
The limit of a finite diagram in exists if and only if the equalizer of two morphisms and exists, where and are constructed using the product structure on (I understand you have already seen this proof somewhere, but just in case Theorem 277). If preserves products, then and should be the and for the diagram composed with , hence taking the equalizer in and applying yields the desired limit because preserves equalizers.
Mike Shulman (Feb 22 2024 at 17:58):I think the point is that if C doesn't have finite products, then and don't exist.
Ralph Sarkis (Feb 22 2024 at 17:59):Oops, I somehow thought you assumed finite products but not all equalisers. (I read "that exist" as if it only applied to equalizers.)
Mike Shulman (Feb 22 2024 at 18:01):What I would try is to write down a very small category that contains a particular finite limit that's not a product or equalizer, like a pullback, and that doesn't have very many products or equalizers. Start with the walking commutative square, perhaps, and maybe add some objects and morphisms if necessary. Then write down an explicit functor out of it (to Set, maybe) that doesn't preserve the pullback, and if the category has few enough products and equalizers you can try to ensure that it preserves them. Ideally, it would have no nontrivial products or equalizers so that last bit would be automatic.
John Baez (Feb 22 2024 at 19:10):This is a case where understanding the word "walking" is very helpful. I would take C to be the walking pullback: the category that contains just enough to contain a pullback square (with no unnecessary equations between objects or morphisms). Similarly I'd take D to be the walking "pullback with a failed contender": the category with just enough to contain a pullback square together with another candidate for being that pullback, which however is not actually the pullback (again with no unnecessary equations between objects or morphisms). . Then let F map the pullback square in C to the failed contender square in D.
Nathan Corbyn (Feb 22 2024 at 19:12):This doesn’t work because the walking pullback has a non-trivial product which must be preserved (it’s not in this example)
John Baez (Feb 22 2024 at 19:14):Oh, whoops, that pullback is a product because one object in the walking pullback is terminal! Okay, so make it not terminal by adding an extra new object with a unique morphism to it from every other.
John Baez (Feb 22 2024 at 19:14):This may take some more finagling... but the plan is to create the "walking counterexample".
John Baez (Feb 22 2024 at 19:16):When you try this sort of argument, sometimes you'll see it fails because you keep needing to add more stuff and you never reach a counterexample. Then, with some luck, you can figure out what's going on and find a proof of the thing you were seeking to disprove!
Nathan Corbyn (Feb 22 2024 at 19:17):I think maybe I posted this question in the wrong channel—if anyone knows the answer I would like to know
Nathan Corbyn (Feb 22 2024 at 19:18):The context is I’m TAing a category theory class at the CS department in Oxford, and one of the exercise sheets made this claim without proof in the statement of one of the exercises. I’m teaching the class tomorrow and I’d prefer to either believe it or have a counter-example to show the class!
John Baez (Feb 22 2024 at 19:19):You posted it in the right channel. You were expecting people to just know the answer instead of trying to figure it out?
Nathan Corbyn (Feb 22 2024 at 19:22):Ah yes sorry, I’d assumed this would be common knowledge
Nathan Corbyn (Feb 22 2024 at 19:24):(As in, I don’t mind not working this one out myself)
John Baez (Feb 22 2024 at 19:25):I think my improved counterexample works, but maybe you can shoot it down.
John Baez (Feb 22 2024 at 19:26):I just took my original attempted counterexample, tacked on a new terminal object in both C and D, and made F preserve that terminal object.
Nathan Corbyn (Feb 22 2024 at 19:28):I think C still has a non-trivial product that isn’t preserved
Nathan Corbyn (Feb 22 2024 at 19:32):I think your original D might be the domain of a counter-example
Nathan Corbyn (Feb 22 2024 at 19:32):The ‘failed contender’ stops the apex of the pullback being a product
Kevin Arlin (Feb 22 2024 at 20:33):Yeah, it's not common knowledge because "a functor preserving all products and equalizers which happen to exist" isn't a natural kind of object. For instance, flat functors are a real thing which reify the intuition of "a functor that wants to preserve all finite limits except maybe they aren't there", but they're a more complicated notion.
Sridhar Ramesh (Feb 24 2024 at 19:34):How's this? Take the free category on objects A, P, B, C, D, with maps from B to D, C to D, P to B, P to C, A to B, and A to C, such that the two composite maps from P to D are equal.
The only parallel non-equal morphisms are the two from A to D, which have no cone and thus no equalizer. Thus, this category has no nontrivial equalizers.
Now consider the unique functor from this to Set which sends P to 0 and all other objects to 1. This preserves all existing products (note that there is no existing product with both B and C as factors). Since there are no nontrivial equalizers, this functor also trivially preserves all equalizers. However, it fails to preserve the pullback square .
Sridhar Ramesh (Feb 24 2024 at 19:39):(This is like John Baez's pullback square + "failed contender", except we do not make the failed contender's square commute, and thus it does not stop the desired pullback from indeed being a pullback.)