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Stream: theory: category theory

Topic: Limits and unnatural transformations

Nathaniel Virgo (Feb 04 2025 at 10:49):

Given functors $F,G: I \to C$, an unnatural transformation $\varepsilon:F\leadsto G$ is a family of maps $\varepsilon_i:F(i)\to G(i)$ for each object $i\in I$, which doesn't have to obey a naturality condition or any other conditions.

Suppose that $F$ and $G$ both have limits, and that we have an unnatural transformation $\varepsilon:F\leadsto G$, together with a cone $K$ over $F$, given by maps $k_i:K\to F(i)$. Suppose also that the maps $k_i{;}\varepsilon_i$ form a cone over $G$ (which is not guaranteed, since $\varepsilon$ isn't natural).

Then we have the unique maps $!_{K,G}:K\to \lim F$ and $!_{K,F}:K\to\lim G$. If $\varepsilon$ were a natural transformation we would also have a map $\lim\varepsilon:\lim F\to \lim G$, but we don't have that in this case.

However, my instinct is that even though we don't have that, $!_{K,G}$ should still factor through $!_{K,F}$. That is, I conjecture that there exists a morphism $\hat\varepsilon:\lim F\to \lim G$ in $C$ (not necessarily unique) such that

$$K\xrightarrow{!_{K,G}} \lim G = K\xrightarrow{!_{K,F}} \lim F\xrightarrow{\hat \varepsilon}\lim G.$$

The morphism $\hat\varepsilon$ will in general depend on the cone $K$ as well as the unnatural transformation $\varepsilon$.

The reason for thinking this is that if we can think of $\lim F$ as "the object of cones over $F$", then the cone $K$ shouldn't "contain any information" that isn't preserved by its map into $\lim F$.

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.

Morgan Rogers (he/him) (Feb 04 2025 at 15:32):

If the $k_i$ happen to be epimorphisms, the condition that the $k_i ; \epsilon_i$ form a cone over $G$ imply naturality of $\epsilon$. Beware that your intuition might be built on examples where this happens to be the case.

Kevin Carlson (Feb 04 2025 at 17:35):

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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 $k.$

Nathaniel Virgo (Mar 06 2025 at 16:15):

Here's another similar conjecture, maybe someone can put me out of my misery again.

Suppose we have functors $F,G:I\to C$ and a natural transformation $\eta:G\Rightarrow F$ where each component has a retraction, i.e. there exists $\varepsilon_i$ such that $\eta_i\mathrel{;}\varepsilon_i = id_{G(i)}$ for each $i\in I$, but the retractions $\varepsilon_i$ only form an unnatural transformation $F\leadsto G$ 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 $\lim \eta:\lim G\to \lim F$ also has a retraction?

These are cofiltered limits if that makes a difference.

Kevin Carlson (Mar 06 2025 at 19:11):

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.

Mike Shulman (Mar 06 2025 at 19:55):

How about the map from the empty set to a nonempty set?

Nathaniel Virgo (Mar 06 2025 at 21:53):

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?

John Baez (Mar 06 2025 at 22:10):

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.

Mike Shulman (Mar 07 2025 at 02:49):

I don't think we have to think about Stone spaces, just consider the category of sets. Let $F: \mathbb{N}^{\rm op} \to \rm Set$ be the functor constant at $\mathbb{N}$, and let $G: \mathbb{N}^{\rm op} \to \rm Set$ be the functor with $G(n) = \{ k\in \mathbb{N} \mid k \ge n\}$. If $n\ge m$ we have $G(n) \subseteq G(m)$, so $G$ is a functor, and $G(n) \subseteq \mathbb{N} = F(n)$ defines a natural transformation $G\Rightarrow F$. Each inclusion $G(n) \hookrightarrow F(n)$ has a retraction, since every monomorphism of sets with nonempty domain has a retraction. But $\lim_n G(n)$ is empty and $\lim_n F(n) = \mathbb{N}$, and the inclusion $\emptyset \hookrightarrow \mathbb{N}$ does not have a retraction.

Kevin Carlson (Mar 07 2025 at 03:29):

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.

Mike Shulman (Mar 07 2025 at 05:01):

What exactly are you saying would be easier?

Kevin Carlson (Mar 07 2025 at 05:26):

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.

Mike Shulman (Mar 07 2025 at 07:58):

Oh. I think infinite sets (especially countably infinite ones) are easier than Stone spaces. (-: