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?

1$\begingroup$ A muddled question with a muddled answer. It would be nice to see an answer that clarifies the meanings of the words algebra (of which there are many) and coalgebra. $\endgroup$– Paul TaylorFeb 13, 2015 at 9:59
1 Answer
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
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 $\beta (f(a)f(x))= \beta (f(ax))$.