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. Peter Johnstone On Thu, 29 Jul 2010, Prof. Peter Johnstone wrote:
Here's a slightly artificial counterexample: let C be the category of finite sets whose cardinality is a power of 2, and all functions between them. The covariant power-set functor restricts to a functor C --> C, and has a monad structure whose algebras are semilattices. If the tensor product of this monad with itself existed, its algebras would be bisemilattices, i.e. sets with two semilattice structures which "commute with each other" in the obvious sense. Free bisemilattices exist, but they don't necessarily have cardinality a power of 2: by my calculation, the free bisemilattice on two generators has seven elements. So the free-bisemilattice functor doesn't exist as an endofunctor of C.
Peter Johnstone ----------------------- On Wed, 28 Jul 2010, Sergey Goncharov wrote:
Dear categorists,
in "Combining algebraic e?ects with continuations", by Hyland et al. the authors say carefully: "In general, the tensor product of two arbitrary monads seems not to exist.." without providing a counterexample though, presumably because they did not have any. Was there any progress reported on this issue since then? Or maybe someone can even make up a counterexample right on the nail?
Thanks,
-- Sergey Goncharov, Junior Researcher
DFKI Bremen Phone: +49-421-218-64276 Safe and Secure Cognitive Systems Fax: +49-421-218-9864276 Cartesium, Enrique-Schmidt-Str. 5 Email: Sergey.Goncharov@dfki.de D-28359 Bremen Site: www.dfki.de/sks/staff/sergey
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