Re: question about monoidal categories

[Jeff Egger <jeffegger@yahoo.ca>]

Hi Paul, 

I think that I have written previously to the list about the 
possibility of a monoidal functor acting on a mere functor, 
and what you have is an instance of this notion.  

Here the monoidal functor is the unique functor M ---> T,
where T is the terminal (monoidal) category.  Your beta is 
a right action of this guy on F.

In general, a right action of monoidal A --U--> C on mere 
P --F--> S requires a right action of A on P and a right 
action of C on S as well as a natural transformation 
  F(p)*U(a) --beta(p,a)--> F(p*a) 
satisfying the appropriate associativity and unit axioms.

In your case S is equipped with the trivial right T-action 
(x*1=x), and M with its canonical right M-action (a*b=a*b).  
The axioms are identical.

Cheers,
Jeff.

--- Paul B Levy <P.B.Levy@cs.bham.ac.uk> wrote:

> 
> 
> 
> Let F be a functor from a monoidal category M to a category S.
> 
> We are given
> 
>          beta(p,a) : F(p) --> F(p*a)
> 
> natural in p,a in M.
> 
> If I tell you that, in addition to naturality, beta is "monoidal", I'm
> sure you will immediately guess what I mean by this, viz.
> 
> (a) for any p,a,b in M
> 
>     beta(p,a) ; beta(p*a,b) = beta(p,a*b) ; F(alpha(p,a,b))
> 
> (b) for any p in M
> 
>     beta(p,1) = F(rho(p))
> 
> Yet I cannot see any reason for giving the name "monoidality" to (a)-(b).
> 
> It doesn't appear to be a monoidal natural transformation in the official
> sense.  There are no monoidal functors in sight.
> 
> Can somebody please justify my usage?
> 
> Paul
> 
> 
> 



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