Morphisms from Hopf algebra to commutative algebra form a group

Question

I have trouble with the following exercise:

Let $H$ be a Hopf algebra with antipode $S$, and let $A$ be a commutative algebra. Show that $$ \mathbf{Alg}(H,A)$$ is a group under convolution with $f^{-1}=f\circ S$.

We have already seen that convolution $*$ is associative and has an identity $\eta_A\circ\varepsilon_H$, so the monoid structure is a given and we only need to show the form of the inverse. I have the following calculation, but nowhere am I using the fact that $A$ is commutative, so there must be a mistake. Where is it?

\begin{align*} f*(f\circ S) &= \mu_A\circ(f\otimes(f\circ S))\circ \Delta\\ &= \mu_A\circ(f\otimes f)\circ (\mathrm{id}_H\otimes S)\circ \Delta \\ &= f\circ \mu_H \circ (\mathrm{id}_H\otimes S)\circ \Delta \\ &= f\circ \eta_H\circ\varepsilon_H \\ &=\eta_A\circ \varepsilon_H \end{align*}

Answer 1

User darij grinberg pointed out in the comments that commutativity is needed to prove the monoid structure. I unfortunately forgot that initially the convolution was defined on the vector space $\mathbf{Vect}(H,A)$, i.e. for linear maps, and closedness was then proved for linear maps only.

Using commutativity, unitality, and associativity we see \begin{align*} (f*g)(hh')&=\sum_{(hh')}f((hh')_{(1)}) g((hh')_{(2)}) \\ &=\sum_{(h)(h')}f(h_{(1)}h'_{(1)}) g(h_{(2)}h'_{(2)}) \\ &=\sum_{(h)(h')}f(h_{(1)})f(h'_{(1)}) g(h_{(2)})g(h'_{(2)}) \\ &=\sum_{(h)(h')}f(h_{(1)})g(h_{(2)})f(h'_{(1)}) g(h'_{(2)}) \\ &=\mu\circ \left((f*g)\otimes (f*g)\right)(h\otimes h') \end{align*} and note also that \begin{align*} (f*g)\circ \eta &= \mu\circ(f\otimes g) \circ \Delta\circ \eta \\ &= \mu\circ(f\otimes g) \circ (\eta\otimes\eta)\\ &= \mu \circ (\eta\otimes\eta) \\ &= \eta\,. \end{align*} Hence $f*g$ is an algebra morphism and the exercise is finished.