Re: Tensor of monads

[Tom Leinster <tl@maths.gla.ac.uk>]

On Fri, 30 Jul 2010, Prof. Peter Johnstone wrote:

> On Fri, 30 Jul 2010, Sergey Goncharov wrote:
>
>>  This looks fine! But I guess, Eckmann-Hilton argument does not apply
>>  to your previous example because it presupposes that the monoidal
>>  structures share the unit, which was not the case there, was it?
>>
> Yes, it was: the fact that the unit for each semilattice structure is
> a homomorphism for the other forces them to be the same.

In any case, it doesn't matter: the Eckmann-Hilton argument *doesn't*
presuppose that the monoid structures share the unit.

Here are the weakest hypotheses I know for the elementary Eckmann-Hilton
argument:

Let A be a set.
Let . be a binary operation on A with two-sided unit 1.
Let * be a binary operation on A with two-sided unit e.
Suppose that

    (a * b) . (a' * b') = (a . a') * (b . b')

for all a, b, a', b' in A.
Then . = *, 1 = e, and (A, ., 1) is a commutative monoid.

So equality of the units and associativity, as well as the more familiar
stuff, come for free.  (I bet you can weaken "two-sided", too.)

Tom


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