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Stream: theory: category theory

Topic: Day convolution

John Baez (Jan 16 2025 at 21:40):

Just a little sanity check. Does all this sound correct? (Has someone bothered to prove it somewhere?)

Suppose $\mathsf{M}$ is a monoidal groupoid and $\mathsf{N}$ is a monoidal subgroupoid, so the inclusion $i : \mathsf{N} \to \mathsf{M}$ is a monoidal functor. For convenience I can assume without loss of generality that $\mathsf{N}$ is [[replete]], so any object of $\mathsf{M}$ isomorphic to an object of $\mathsf{N}$ is already in $\mathsf{N}$.

Then restriction of presheaves from $\mathsf{M}$ to $\mathsf{N}$ gives a functor

$i^\ast: \mathsf{Set}^{M^{\text{op}}} \to \mathsf{Set}^{N^{\text{op}}}$

that is symmetric monoidal with respect to Day convolution. Its left adjoint

$i_!: \mathsf{Set}^{N^{\text{op}}} \to \mathsf{Set}^{M^{\text{op}}}$

is "extension of presheaves from $\mathsf{N}$ to $\mathsf{M}$, defining them to be $0$ on objects not in $\mathsf{N}$", and this is also symmetric monoidal. Furthermore,

$i_! i^\ast \cong 1$

as symmetric monoidal functors.

John Baez (Jan 16 2025 at 21:43):

(I haven't bothered to think about the case where we have categories rather than groupoids.)

Sam Staton (Jan 17 2025 at 09:37):

Hi John, I think the left adjoint $i_!$ being monoidal follows from the universal property (Im and Kelly 1986)?
But are you sure about $i^*$? I don't think it is strong monoidal in the case
$i : \mathbf{FinSetBij}^{\mathrm{op}}\to \mathbf{FinSetFun}^{\mathrm{op}}$,
considered with the disjoint union monoidal structure. But you said groupoid, and maybe there is a trick.

Sam Staton (Jan 17 2025 at 09:42):

Oh, but sorry maybe you didn't mean strong monoidal. In that case I think the right adjoint $i^*$ being monoidal is remarked in Day and Street, Kan extensions along promonoidal functors, 1995.

John Baez (Jan 17 2025 at 18:45):

Thanks very much! I actually meant strong monoidal, but my guiding examples have a feature I failed to mention here: the monoidal subgroupoid $i : \mathsf{N} \to \mathsf{M}$ is a full subcategory. You're making me think this may be crucial. I'll think about it more.

A full monoidal subgroupoid of a monoidal groupoid $\mathsf{M}$ is a rather simple thing: to specify it we just pick a submonoid of the monoid of connected components $\pi_0(\mathsf{M})$.

Chaitanya Leena Subramaniam (Jan 17 2025 at 22:21):

I think everything you wrote is correct (modulo the remark below) when $N\subset M$ is a full monoidal embedding of categories with biproducts (assuming the quick co-end calculus I just did doesn't have a mistake). Otherwise, I'm not sure how to show that $i^*$ is strong monoidal.

Remark: Also, are you sure you don't mean that $i^*i_!\cong 1$? For this is equivalent to $i_!$ being fully faithful, which in turn is equivalent to $i$ being fully faithful.

John Baez (Jan 17 2025 at 22:56):

Thanks! Yes, I should have said $i^* i_! \cong 1$.

Kevin Carlson (Jan 17 2025 at 23:26):

I thought this was the quantum mechanics thread still briefly and starting squinting real hard at "$i^*i_!$".