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Stream: theory: category theory

Topic: reflecting structure

Matteo Capucci (he/him) (Aug 25 2023 at 09:51):

Categories-with-structure can often be thought as pseudoalgebras of some 2-monad on $\bf Cat$, whose maps preserve their structure (up to coherent iso). I wonder how to make sense of maps reflecting structure instead. Any idea?

James Deikun (Aug 25 2023 at 10:25):

I don't have an answer to this precise question off the top of my head, but there's a whole network of preservation theorems similar to the HSP theorem for different logical doctrines and some of them involve reflection of structure. Functors reflecting structure probably involve a categorified version of one of these doctrines.

Ralph Sarkis (Aug 25 2023 at 14:54):

What does it mean to reflect structure in general? I know about order-reflecting, but I can't wrap my head around what it would mean for a function to reflect a binary operation (say, of a monoid).

Matteo Capucci (he/him) (Aug 25 2023 at 15:35):

Well uhm good question... In general it seems to mean that if C and D both have Xs, then $F:C \to D$ reflects them when an X for F(c) in D is actually given by F(c'), where c' is X of c in C.

So for a monoidal structure... I guess it just means preserving? Because there is at most one $c_1 \otimes c_2$ and thus we are asking for $F(c_1)\otimes F(c_2)$ to be $F(c_1 \otimes c_2)$.

Also in general one talks about reflecting structure between categories which might not have all that structure... So you can have a limit-reflecting functor landing in a category with not so many limits.

Mike Shulman (Aug 25 2023 at 20:26):

For sets with structure, we can say that a function $f:X\to Y$ preserves some $n$-ary relations $R$ and $S$ when $f_!(R) \subseteq S$, or equivalently $R \subseteq f^*(S)$. If $R$ and $S$ are graphs of $(n-1)$-ary operations, that's are equivalent to saying that $f$ commutes with those operations.

Dually, we could say that it reflects it if $f^*(S) \subseteq R$, or equivalently $S\subseteq f_*(R)$. In the graph case I think that's a bit weaker than saying that $f$ commutes with the operations; in the case of a binary operation it says that if $f(x) \otimes f(y) = f(z)$ then $z = x \otimes y$, but doesn't assert that the image of $f$ is closed under the operation.

To categorify that, I guess we'd have to replace relations with spans or fibrations or something.