A morphism of bialgebras between two Hopf algebras is necessarily a morphism of Hopf Algebras.

Question

Here is the question I am trying to solve:

Use the previous exercise to show that a morphism of bialgebras between two Hopf algebras is necessarily a morphism of Hopf algebras.

Here is the previous exercise:

(Convolution and Composition) Consider a morphism of algebras $f: A \to A'$ and a morphism of coalgebras $g: C' \to C.$ Prove that the map $h \to f \circ h \circ g$ from $\operatorname{Hom}(C, A)$ to $\operatorname{Hom}(C', A')$ is a morphism of algebras for the convolution $\star.$

I got a hint to prove that $ S\circ f = f \circ S $ by applying left and right convolution with $f = f \circ id = id \circ f,$ where $S$ is an antipode of a bialgebra $H$ which is an endomorphism of $H$ that satisfies $$S \star id_H = id_H \star S = \eta \circ \varepsilon . $$

I still do not know how we should the previous exercise in the solution. Could anyone explain this to me please?

Also, I saw a solution for this problem here Definition morphism of Hopf algebra but I am not sure

And I am not sure what is the proof of uniqueness of inverse in the convolution algebra, could anyone show me the proof please? Also, is there an implicit assumption that we have a commutative algebra?

Also, if this solution uses the problem I mentioned before ( previous exercise), which step exactly did it use it?

Answer 1

Let $H$ and $H'$ be Hopf algebras. You know $\operatorname{Hom}(H, H')$ is an associative algebra with multiplicative unit $\eta_{H'} \circ \epsilon_H$, where $\epsilon_H$ denotes the counit map of $H$ and $\eta_{H'}$ denotes the unit map of $H'$.

In the linked post, given a bialgebra morphism $f \in \operatorname{Hom}(H, H')$ it is proved using the definition of antipode that the following two identities hold:

1. $(f \circ S_H) \star f = \eta_{H'} \circ \epsilon_H.$
2. $f \star (S_{H'} \circ f) = \eta_{H'} \circ \epsilon_H.$

Now you can conclude that $f \circ S_H = S_{H'} \circ f$ as follows. Let us denote $\eta_{H'} \circ \epsilon_H$ by $\pmb{1}$. Then \begin{align} f \circ S_H = (f \circ S_H)\star \pmb{1} &= (f \circ S_H) \star \Big( f \star (S_{H'} \circ f) \Big) \\ &= \Big( (f \circ S_H) \star f \Big) \star (S_{H'} \circ f) \\ &= \pmb{1} \star (S_{H'} \circ f) \\ & = S_{H'} \circ f. \end{align} You may have seen this type of argument in the monoids or something similar.