On Fri, 30 Jul 2010, Sergey Goncharov wrote:
Sorry, the previous posting was nonsense -- a bisemilattice is the same thing as a semilattice, by the Eckmann-Hilton argument. However, if you leave out the zero, and consider the "set of nonempty subsets" monad, this time on the category of sets of cardinality 2^n - 1 for some n, you do get a counterexample. This looks fine! But I guess, Eckmann-Hilton argument does not apply to your
On 07/29/2010 03:24 PM, Prof. Peter Johnstone wrote: 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. Peter Johnstone P.S. -- You can enlarge the base category to contain all finite sets of odd cardinality (so that, for example, it's cartesian closed). The free bi-(semilattice-without-unit) on three generators has 20 elements. [For admin and other information see: http://www.mta.ca/~cat-dist/ ]