Finite dual of an algebra morphism. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:34:01Z http://mathoverflow.net/feeds/question/44738 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44738/finite-dual-of-an-algebra-morphism Finite dual of an algebra morphism. Neha 2010-11-03T22:09:20Z 2010-11-04T09:48:20Z <p>Is finite dual of an algebra morphism a morphism of coalgebras? Does taking finite dual preserves exactness of an exact sequence of algebra morphisms? When is this possible?</p> http://mathoverflow.net/questions/44738/finite-dual-of-an-algebra-morphism/44795#44795 Answer by Bugs Bunny for Finite dual of an algebra morphism. Bugs Bunny 2010-11-04T09:48:20Z 2010-11-04T09:48:20Z <p>Gee, I have not a clue what is up, doc! I have never seen an exact sequence of algebra morphisms because usual linear kernels are not subalgebras. Having said that, algebra morphisms may conceivably have kernels but you need to expand on that</p> <p>BTW, the answer to the first question is yes! All it is kinda saying that $^0$ is a functor. Say $f:A->B$ is an algebra morphism. The equality $\sum_{(\beta)} f^0(\beta_1)\otimes f^0(\beta_2)= \sum_{(f^0(\beta))} f^0(\beta)_1\otimes f^0(\beta)_2$ can be checked on any pair of elements $a,x\in A$ where it becomes $\beta (f(a)f(x))= \beta (f(ax))$.</p>