From: Fred Linton
I was relying on the Axiom of Choice (AC) to know that every vector space (scalars from an arbitrary field) has a basis. Paraphrasing: every vector space is free. Thinking (loosely) of the Kleisli category as the full subcategory of the EM category consisting of the free algebras, this means the K and the EM categories "coincide" (to within equivalence of categories). The exact nature of the field is irrelevant. But AC (or something else strong enough to guarantee every vector space has a basis) is crucial.
This seems right for ordered fields. Is it also true for finite fields and other non-ordered fields like the complex numbers? If every Archimedean field k gives rise to a Klem (Kleisli ~ EM) monad Vct_k, that's uncountably many Klem monads right there, all in the lattice of submonads of Vct_R. (The submonads of a monad must form a complete lattice, more precisely a complete semilattice, if they're closed under arbitrary intersection; is it always an algebraic semilattice, as for algebras?) k doesn't even need to be a field or an additive group, any subrig of R will do, as Bill was hinting at. (The coefficients have to form a rig for T to be functorial.) Call this Method A for forming a submonad of Vct_R. Method B is to limit the operations to those whose weight (sum of coefficients) is drawn from a submonoid of the monoid R under multiplication. (It has to be a submonoid for the multiplication mu and unit eta to remain defined.) The smallest such monoid is {1}, which gave rise to the affine spaces as per my previous post. R has plenty of other multiplicative submonoids, such as the set {c^i} for any real c > 0 and i ranging over any submonoid of Z under addition, or for any real c and i ranging over any submonoid of N under addition. This can be combined with Method A (coefficients from a subrig of R) to get even more submonads. Questions: 1. Do all submonads of Vct_R arise as above (coefficients limited to a subrig of R, weights limited to a multiplicative submonoid of R)? 2. Among the submonads of Vct_R, are the Klem ones exactly those for which the set of coefficients is dense in some open interval of R?
But not for 1+1+X: the free bipointed sets are in a natural bijection with the non-free ones, the latter having the two points identified
I'm not entirely sure what's "natural" here. Certainly the free bipointed sets are in bijective correspondence with the (just plain) sets: that's a "consequence," if you like, of the canonical functor from SETS to any Kleisli category over SETS being a bijection on objects. So the non-free bipointed sets, coinciding as they do with the (singly) pointed sets, are likewise in bijective correspondence with the (just plain) sets. But the same is true for the free algebras over ANY monad on sets. So? "This"?
... --this must be about the simplest nontrivial instance of obtaining an algebra as a quotient of a free algebra, not a bad introductory example when explaining how to get algebras from free algebras.
Point taken. I was trying to say that identifying the two constants of a free bipointed set was the canonical way of producing any nonfree bipointed set. What's a slicker way to say this? Vaughan