Subject: Re: pullback categories

From: categories <[email-masked]>
Date: Tue, 20 Jan 1998 15:02:18 -0400 (AST)

Date: Tue, 20 Jan 1998 14:20:35 -0500 (EST)
From: Michael Barr <[email-masked]>

 Triples is back, at least for the time being, after having been down,
 either from its problems or the lack of power at McGill for most of the
 last month.
 
 I have just seen the question about pullback of tripleable functors and
 I have not seen a really satisfactory reply.  A lot--maybe all--of what
 I say below is probably in Ernie Manes' thesis (A Triple Miscellany,
 Wesleyan U., 1967).  The answer is definitely no, but the problem is the
 lack of adjoint.  Consider, for the example, the complete semilattice
 triple on Set.  It is also the covariant powerset triple, with singleton
 for eta and union (or intersection) for mu (one will give you sup
 semilattices, the other the inf semilattices).  That is one triple and
 the other is N x -, whose algebras are sets with a single unary
 operation.  The pullback is the category whose objects are complete
 semilattices with a single unary function that is not assumed to cohere
 in any way with the semilattice structure.  A complete boolean algebra
 is model of such a theory, taking complement as the unary operation.  If
 there were free algebras of this type, then a quotient of them would be
 a free complete boolean algebra, but we know these don't exist.
 
 Of course, one can raise the question under the assumption that the
 adjoint exists.  For example, if the category is locally presentable
 (complete and accessible) and the functors are accessible.  Then if the
 functors are T_1 and T_2, simply apply them alternately, taking colimits
 at limit ordinals, until they stabilize and that will give you free
 algebras.  Of course, it is obvious that the pullback is the category
 whose objects consist of T_1A ---> A <--- T_2A which are algebras for
 each triple, but no assumption of coherence between the two structures
 is made.  It seems pretty obvious, although I have no really checked the
 details carefully, that this will satisfy Beck's condition.  Or rather
 the forgetful functor to each of the individual category of algebras as
 well as the composite.  For example, if we had the situation
            T_1A -----------> A <------------- T_2A
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             vv               vv                vv
            T_1B -----------> B <------------- T_2B
 of such a nature that A ======> B -----> C is a split coequalizer, then
 the outer columns of
            T_1A -----------> A <------------- T_2A
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             ||               ||                ||
             vv               vv                vv
            T_1B -----------> B <------------- T_2B
             |                |                 |
             |                |                 |
             |                |                 |
             |                |                 |
             |                |                 |
             |                |                 |
             v                v                 v
            T_1C - - - - - -> C <- - - - - - - T_2C
 are split coequalizers too, whence the dotted arrows exist.  That the
 whole diagram is a coequalizer in the pullback category is also easy.
 Thus the answer is yes, provided the adjoint exists, but that is not
 guaranteed even over Set.
 
 Michael