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Stream: learning: questions

Topic: adjoints to discrete monoidal categories

Emily (Sep 19 2021 at 20:42):

Does the functor
$(-)_\mathsf{disc}\colon\mathsf{Mon}\to\mathsf{MonCats}^{\mathsf{strong}\text{ }\mathsf{monoidal}\text{ }\mathsf{functors}}$
sending a monoid to its associated discrete monoidal category have a left or right adjoint?

Emily (Sep 19 2021 at 20:42):

I think it does by the pseudomonoid version of this together with the adjunction $\pi_0\dashv(-)_\mathsf{disc}\dashv\mathrm{Obj}\dashv(-)_\mathsf{indisc}$, but I haven't checked it yet..

sarahzrf (Sep 19 2021 at 21:05):

hmm... i would be inclined to guess that one of the adjoints is some kind of grothendieck group–like construction

sarahzrf (Sep 19 2021 at 21:08):

I'm pretty sure the left adjoint is gonna be "connected components of the category, equipped w/ the monoid structure [A][B] = [A ⊗ B]"

sarahzrf (Sep 19 2021 at 21:08):

(where [A] is the connected component of A)

Emily (Sep 19 2021 at 21:31):

Thanks, Sarah! I think the pseudomonoid argument should give precisely this description, but at the same time I'm a bit confused. Firstly, $\pi_0$ is indeed strong monoidal, with $\pi_0(A\times B)\cong\pi_0(A)\times\pi_0(B)$ and $\pi_0(\mathsf{pt})\cong\mathrm{pt}$, as is $(-)_\mathsf{disc}$, so we should get precisely your construction.

OTOH, each of the other functors is also strong monoidal, and so the pseudomonoid nonsense would (?) seem to say that the right adjoint to disc is Obj again, but I can't see how to put a monoid structure on it without first passing to isomorphism classes... (I think there's an error in this last part (that we get Obj again). I have to go now, but I'll try to figure this out in a bit. Thanks!)

Emily (Sep 19 2021 at 21:38):

Actually let me just mention this too before going: https://mathoverflow.net/questions/122105

one does get precisely your construction :)