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Stream: learning: questions

Topic: reverse of a monoidal category?

Martti Karvonen (Mar 22 2022 at 13:10):

Is there a standard name for the "reverse" of a monoidal category, where you keep the underlying category the same but switch the order for the monoidal product (so that $A\otimes_{new} B:=B\otimes A$)?

Reid Barton (Mar 22 2022 at 13:17):

FWIW (not much) I also call it the "reverse"

JS PL (he/him) (Mar 22 2022 at 13:26):

I was going to suggest looking at what the terminology is for monoids/groups/rings, where $x \ast_{new} y = y \ast x$, but it's called the opposite monoid/group/ring.

Oscar Cunningham (Mar 22 2022 at 13:28):

If you treat it as a 2-category on one object then it's also called the opposite: https://ncatlab.org/nlab/show/opposite+2-category

JS PL (he/him) (Mar 22 2022 at 13:33):

An argument could be made that you can then call this the opposite monoidal category, which says that the opposite is on the monoidal structure, vs monoidal opposite category, which I guess says the opposite is on the underlying category.

Reid Barton (Mar 22 2022 at 13:34):

I think a better argument would be that you should use a different word altogether, like "reverse" :upside_down:

Reid Barton (Mar 22 2022 at 13:35):

For this kind of reason I think it's not actually a good idea to simply regard a monoidal category as "being" a one-object 2-category. There is a correspondence between them but it changes the meaning of op, co, etc

JS PL (he/him) (Mar 22 2022 at 13:37):

Reid Barton said:

I think a better argument would be that you should use a different word altogether, like "reverse" :upside_down:

Oh absolutely!

Joe Moeller (Mar 22 2022 at 18:49):

I've seen it called the reverse. I think that's what they say in EGNO "Tensor Categories".

Matteo Capucci (he/him) (Mar 23 2022 at 08:42):

Also Johnson and Yau call this 'reverse' in Example 1.2.9 of their book on 2-dimensional category theory

Matteo Capucci (he/him) (Mar 23 2022 at 08:43):

Then one hass $B(\mathcal M^{rev}) = (B\mathcal M)^{op}$, while $B(\mathcal M^{op}) = B\mathcal M^{co}$

Mike Shulman (Mar 23 2022 at 15:31):

I think "reverse" is pretty common.

John Baez (Mar 24 2022 at 00:09):

Oh-oh - it seems like everyone is agreeing on this terminology. What's wrong?