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