Can we construct free semi-lattice from free dist. lattice?

Can we construct free semi-lattice from free dist. lattice?

[Christopher Townsend]

Dear All,
I have a problem which I had thought very specialised, but actually can be
stated very generally: -

I have a category C with finite limits, and so I also have a category
DLat(C) of distributive lattices which, lets say, has coequalizers. If free
distibrutive lattices can be constructed (i.e. if there exists F:C->DLat(C)
left adjoint to the forgetful functor) then do free semilattices exist?

Thanks, Christopher Townsend (OU)

Re: Can we construct free semi-lattice from free dist. lattice?

[Prof. Peter Johnstone]

On Tue, 27 May 2003, Christopher Townsend wrote:

> Dear All,
> I have a problem which I had thought very specialised, but actually can be
> stated very generally: -
>
> I have a category C with finite limits, and so I also have a category
> DLat(C) of distributive lattices which, lets say, has coequalizers. If free
> distibrutive lattices can be constructed (i.e. if there exists F:C->DLat(C)
> left adjoint to the forgetful functor) then do free semilattices exist?
>
I suspect the answer is no, for irritatingly trivial reasons.
If C is a pointed category (i.e. has a zero object), then any internal
distributive lattice in C has its top and bottom elements equal, and
so is degenerate (i.e. isomorphic to the terminal object 1). Hence
the free-distributive-lattice functor exists, and is the constant
functor with value 1. But I'm sure there must be examples of pointed,
finitely complete categories which don't have a free-semilattice
functor (though I have to admit I don't have one at my fingertips).

Peter Johnstone