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Given functors , an unnatural transformation is a family of maps for each object , which doesn't have to obey a naturality condition or any other conditions.
Suppose that and both have limits, and that we have an unnatural transformation , together with a cone over , given by maps . Suppose also that the maps form a cone over (which is not guaranteed, since isn't natural).
Then we have the unique maps and . If were a natural transformation we would also have a map , but we don't have that in this case.
However, my instinct is that even though we don't have that, should still factor through . That is, I conjecture that there exists a morphism in (not necessarily unique) such that
The morphism will in general depend on the cone as well as the unnatural transformation .
The reason for thinking this is that if we can think of as "the object of cones over ", then the cone shouldn't "contain any information" that isn't preserved by its map into .
Although this seems kind of plausible to me I don't see a way to prove it, so my question is whether it's true, or if it's not true in general, what other assumptions might need to be in place to make it so.
If the happen to be epimorphisms, the condition that the form a cone over imply naturality of . Beware that your intuition might be built on examples where this happens to be the case.
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Here’s the minimal counterexample I can think of: the universal category containing two cospans, cones over those cospans, and an unnatural transformation between them, adjoined with an initial object. The two cones summitting the filled squares are indeed pullbacks since the only other cones over the cospans are those under the initial object
Here's another similar conjecture, maybe someone can put me out of my misery again.
Suppose we have functors and a natural transformation where each component has a retraction, i.e. there exists such that for each , but the retractions only form an unnatural transformation and not a natural one. (Retractions aren't unique of course - I don't mind whether we consider this as a specified family of retractions or whether we just say the retractions exist.)
Then can we expect that also has a retraction?
These are cofiltered limits if that makes a difference.
I’m wondering whether you can find a non-split mono of Stone spaces. I can’t think of one now and perhaps they don’t even exist but that should resolve it.
How about the map from the empty set to a nonempty set?
Does that mean there's a counterexample? Is there an easy to way to see it for someone who doesn't know anything about Stone spaces?
I don't know what's going on in this discussion, but the category of Stone spaces is just the opposite of the category of boolean algebras, if that helps. Mike's nasty counterexample translates into: there's an epimorphism from the 2-element boolean algebra to the 1-element boolean algebra that doesn't split. The reason is that in the 1-element boolean algebra we have T = F, and to map this algebra to the 2-element boolean algebra we'd have to map this element to two different things, T and F.
I don't think we have to think about Stone spaces, just consider the category of sets. Let be the functor constant at , and let be the functor with . If we have , so is a functor, and defines a natural transformation . Each inclusion has a retraction, since every monomorphism of sets with nonempty domain has a retraction. But is empty and , and the inclusion does not have a retraction.
Indeed you’ve made it just a bit harder than working with Stone spaces, since if those functors had been finite set valued, Stone spaces is exactly what you’d have! I think the infinite values were necessary to get an empty limit, though.
What exactly are you saying would be easier?
All I meant is it would’ve been easier if it were possible to do this with diagrams of finite sets because finite sets are easier than infinite ones.
Oh. I think infinite sets (especially countably infinite ones) are easier than Stone spaces. (-: