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?
closed as offtopic by Ricardo Andrade, Neil Strickland, quid, Johannes Hahn, Peter Michor Feb 14 at 19:27This question appears to be offtopic. The users who voted to close gave this specific reason:



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 $\beta (f(a)f(x))= \beta (f(ax))$. 

