Re: Q about_monoidal_functors?

[Toby Bartels <toby+categories@ugcs.caltech.edu>]

Andre Joyal wrote in part:

>We should use a similar terminology for spaces and maps.
>E-n space <--> E-n map
>Also for (higher) categories and functors.
>monoidal category <---> monoidal functor
>braided monoidal category <----> braided monoidal functor
>2-braided monoidal category <--> 2-braided monoidal functor
>3-braided monoidal category <--> 3-braided monoidal functor
>......
>symmetric monoidal category <--> symmetric monoidal functor

I agree, one should say "symmetric monoidal functor";
if nothing else, that indicates that the source and target
are symmetric (not merely braided) monoidal categories.
I only put "BMF" in my table to show a particular pattern.

Depending on how you write down the definitions,
that a braided monoidal functor between symmetric monoidal categories
is the same thing as a symmetric monoidal functor between them
is either an utter triviality or a deep and interesting theorem;
but in either case, we need the words to state it.

(I do agree with John about preferring "k-tuply monoidal",
  but I'll let him make that argument.)

>A (n+1)-braided monoidal n-category is symmetric by
>the stabilisation hypothesis.
>I believe that a (n+1)-braided monoidal functor
>between (n+1)-braided monoidal n-categories is symmetric.

I think that you mean to say (which is even stronger)
that an n-braided monoidal functor between SM n-categories is symmetric.
More generally, a k-braided monoidal l-transfor between SM n-categories
is symmetric as long as k + l is greater than or equal to n.
(A 0-transfor is a functor, a 1-transfor is a natural transformation, etc.
  This numbering is due to Sjoerd Crans; feel free to argue that it's off.)

More generally yet, a k-braided monoidal l-transfor
between m-braided monoidal n-categories is m-braided,
as long as k + l >= n, regardless of the value of m
(although we need m >= k for the antecedent to make sense).

>Is this part of the official stabilisation hypothesis?

I don't know what's official, but I'll claim the conjecture above as mine
if nobody else has written it down yet.  (^_^)


--Toby


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