# Finite dual of an algebra morphism. [closed]

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 off-topic by Ricardo Andrade, Neil Strickland, quid, Johannes Hahn, Peter MichorFeb 14 '15 at 19:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Ricardo Andrade, Community, Johannes Hahn, Peter Michor
If this question can be reworded to fit the rules in the help center, please edit the question.

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. – Paul Taylor Feb 13 '15 at 9:59

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