Skip to main content
2 of 3
added 609 characters in body
Todd Trimble
  • 53.3k
  • 6
  • 205
  • 322

Under the most immediate reading of "both of type A and B", where no sort of compatibility or relationship between the A and B structures is assumed, the answer is no.

A counterexample is where A is the theory of sup-lattices and B is the theory of sets equipped with an unary operator (often called "successor"). If free algebras of the combined theory existed, then in particular there would exist the initial such algebra = free algebra on 0 generators. This can be seen as an algebraic form of the cumulative hierarchy, where such a structure is provably isomorphic to the set of all its small subsets, in contradiction to Cantor's theorem. In other words, there is no such small structure.

In fact, these observations are the beginning of Algebraic Set Theory. See the text by Joyal and Moerdijk (London Math. Soc. LNS 220), or for a very accessible introduction, try these notes by Steve Awodey, where the inexistence is proved on page 2.

Edit: But it seems that I should at least add that if theories A and B have rank (are given by small sets of operation symbols and equational axioms), then the combined theory will also have rank and hence will admit free algebras. In the finitary case, Lawvere indicated in his 1963 thesis that the category of what are now called Lawvere theories admits coproducts, and infinitary Lawvere theories with rank also admit coproducts, which represent the combined theory. It would take a little while however to describe the construction of coproducts here. If you are interested I could say more.

Todd Trimble
  • 53.3k
  • 6
  • 205
  • 322