The best answer was that of Ernie Manes. What is at stake is one very simple observation: if A is the category of algebras for a monad (perhaps in the context of enriched category theory) and F is the full subcategory given by the free algebras, then F is dense in A, and moreover this density may be "presented" by coequalizers. For the meaning of "density presentation", a notion due to Brian Day, see p.172 of my book "Basic Concepts of Enriched Category Theory", CUP 1982. Here it means that every algebra is a coequalizer of a pair of maps between free algebras, IN SUCH A WAY THAT this colimit is preserved by the representables A(f,-) where f is a free algebra; and this is very easy to see. Now Thm 5.30 of my book gives a very simple proof that the left Kan extension along J: F --> A of any functor T: F --> B exists, provided only that B has coequalizers. The observation of Dusko Pavlovic, that an elegant argument is to hand when B too is monadic, gives too SPECIAL a result. The others that have answered don't seem to have pointed out that the "canonical" extension of T they refer to IS indeed the left Kan extension. Max Kelly.