Looking at the two ways of evaluating (say with . vertically and * horizontally) 1 e e 1 is what gives 1=e but it seems to need two sided units. Ronnie Tom Leinster wrote:
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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