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From: "Prof. Peter Johnstone"
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Subject: Re: Tensor of monads
Date: Thu, 29 Jul 2010 14:24:43 +0100 (BST)
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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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