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Stream: theory: category theory

Topic: Double-gluing and functors

Aleks Kissinger (Apr 26 2020 at 12:59):

Question: Suppose I have a *-preserving strong symmetric monoidal functor between *-autonomous categories $F: \mathcal C \to \mathcal D$. Can that be lifted (possibly under some assumptions) to the same kind of functor between categories $G(\mathcal C) \to G(\mathcal D)$ arising from a double-gluing construction?

To use the notation of @Andrea Schalk and Hyland's paper https://core.ac.uk/download/pdf/21173316.pdf (p.32):
I am especially interested in the case of the tight categories $T(\mathcal C), T(\mathcal D)$ arising from double-gluing with a focus $F := \{ 1_I \}$, where $I$ is the unit of both tensor and par (i.e. an ISOMIX category).

Mike Shulman (Apr 27 2020 at 23:08):

Well, the $\ast$-autonomous structure of double-glued categories is constructed from the analogous structure of the input categories plus limits. So I would expect that if you have a functor that preserves all that structure, including the relevant limits, then it should induce a structure-preserving functor on the double-glued categories.

Aleks Kissinger (Apr 28 2020 at 09:15):

Hmm. I haven't spent much time thinking about the general construction (as opposed to the concrete hom-functor case). Here, I have two categories glued over the same target. So, construct $G(\mathcal C)$ from $L : \mathcal C \to \mathcal E$ and $G(\mathcal D)$ from $L' : \mathcal D \to \mathcal E$. So, in addition to $F : \mathcal C \to \mathcal D$ preserving *-autonomous structure, I'll probably want $L = L' F$. In the case where $L$ and $L'$ are $hom(I, -)$, this looks fine. For the multiplicative structure from double-gluing, $\mathcal E$ needs pullbacks, but $\mathcal C, \mathcal D$ don't need any limits, so there's nothing else obvious for $F$ to preserve.

I guess it's probably time to shut up and calculate. :)

Andrea Schalk (May 06 2020 at 13:25):

Is this question still open? For the first case you give, to define the functor you need to make a choice, I think. Assume you have $(R,U,X)$ in the tight orthogonality category for $\mathcal{C}$. While it is clear that $(FR,F[U],F[X])$ will be an object in the slack orthogonality category for $\mathcal{D}$, in order to ensure you obtain something in the tight subcategory you either have to ask for extra properties for $F$, or you have to apply a closure operation, but then you have to make a choice between using $(FR,(F[U])^{00},(F[U])^0)$ and $(FR,(F[X])^0,(F[X])^{00})$.

(=_=) (May 06 2020 at 14:42):

Andrea Schalk said:

Assume you have $(R,U,X)$ in the tight orthogonality category for $$\cal{C}$$. While it is clear that $(FR,F[U],F[X])$ will be an object in the slack orthogonality category for $$\cal{D}$$,

"\cal" is not a valid command. You're probably looking for "\mathcal", i.e.

Assume you have $(R,U,X)$ in the tight orthogonality category for $\mathcal{C}$. While it is clear that $(FR,F[U],F[X])$ will be an object in the slack orthogonality category for $\mathcal{D}$,

Aleks Kissinger (May 12 2020 at 11:05):

@Andrea Schalk Ah yes, that makes sense. I guess an extra property for $F$ that would work is $F[U^0] = F[U]^0$, but this seems very strong. Do you think anything weaker would do the job?