Uniqueness and anti-multiplicativity and anti-comultiplicativity of the antipode

If $p,q$ are two two-sided convolution inverses of the same map $r$ in a unital associative convolution algebra with unit $e$, then $p=p\ast (r\ast q)=(p\ast r)\ast q=e\ast q=q$. Taking $r={\mathrm{i}\mathrm{d}}_H$ proves uniqueness of the antipode.

Give ${\mathrm{Hom}}_k(H\otimes H,H)$ convolution using the coalgebra of [F2] and the algebra of $H$. Define $\varphi =S\circ m$ and $\psi =m\circ (S\otimes S)\circ \tau$, which are well-defined by [F6]; thus $\psi (a\otimes b)=S(b)S(a)$. Both are two-sided inverses of $m$: $(\varphi \ast m)(a\otimes b)=\sum S(a_{(1)}{b}_{(1)})a_{(2)}{b}_{(2)}=\epsilon (a)\epsilon (b)1$ and $(m\ast \varphi )(a\otimes b)=\sum a_{(1)}{b}_{(1)}S(a_{(2)}{b}_{(2)})=\epsilon (a)\epsilon (b)1$ by the antipode equations at $ab$, multiplicativity of $\mathrm{\Delta }$, and multiplicativity of $\epsilon$; also $(\psi \ast m)(a\otimes b)=\sum S(b_{(1)})S(a_{(1)})a_{(2)}{b}_{(2)}=\epsilon (a)\epsilon (b)1$ and $(m\ast \psi )(a\otimes b)=\sum a_{(1)}{b}_{(1)}S(b_{(2)})S(a_{(2)})=\epsilon (a)\epsilon (b)1$ by first contracting $S(a_{(1)})a_{(2)}$ or $b_{(1)}S(b_{(2)})$ and then the remaining antipode pair. The right side is the convolution unit on $H\otimes H$ by [F2].

Give ${\mathrm{Hom}}_k(H,H\otimes H)$ convolution using $\mathrm{\Delta }$ on the source and the tensor-product algebra on the target. The map $\mathrm{\Phi }=\mathrm{\Delta }\circ S$ is a two-sided inverse of $\mathrm{\Delta }$: $(\mathrm{\Phi }\ast \mathrm{\Delta })(h)=\sum \mathrm{\Delta }(S(h_{(1)}))\mathrm{\Delta }(h_{(2)})=\mathrm{\Delta }(\sum S(h_{(1)})h_{(2)})=\epsilon (h)(1\otimes 1)$ and $(\mathrm{\Delta }\ast \mathrm{\Phi })(h)=\sum \mathrm{\Delta }(h_{(1)})\mathrm{\Delta }(S(h_{(2)}))=\mathrm{\Delta }(\sum h_{(1)}S(h_{(2)}))=\epsilon (h)(1\otimes 1)$ by multiplicativity of $\mathrm{\Delta }$ and the two antipode equations. Here the convolution unit is $h\mapsto \epsilon (h)(1\otimes 1)$.

Define $\mathrm{\Psi }:=(S\otimes S)\tau \mathrm{\Delta }:H\to H\otimes H$. Parenthesize the coassociative fourfold coproduct using [F7] as $\sum h_{(1)}\otimes (h_{(2)}{)}_{(1)}\otimes (h_{(2)}{)}_{(2)}\otimes h_{(3)}$. In $\mathrm{\Psi }\ast \mathrm{\Delta }$, the adjacent middle product is $S((h_{(2)}{)}_{(1)})(h_{(2)}{)}_{(2)}$, so the left antipode equation and then the counit equation give $(\mathrm{\Psi }\ast \mathrm{\Delta })(h)=\sum S(h_{(2)})h_{(3)}\otimes S(h_{(1)})h_{(4)}=\sum 1\otimes S(h_{(1)})h_{(2)}=\epsilon (h)(1\otimes 1)$. In $\mathrm{\Delta }\ast \mathrm{\Psi }$, the adjacent middle product is $(h_{(2)}{)}_{(1)}S((h_{(2)}{)}_{(2)})$, so the right antipode equation and then the counit equation give $(\mathrm{\Delta }\ast \mathrm{\Psi })(h)=\sum h_{(1)}S(h_{(4)})\otimes h_{(2)}S(h_{(3)})=\sum h_{(1)}S(h_{(2)})\otimes 1=\epsilon (h)(1\otimes 1)$. The tensor maps are well-defined by [F6], and coassociativity makes these contractions independent of the chosen Sweedler expression.

Two-sided convolution inverses are unique by the calculation in step 1.1, applied with common map $m$ in ${\mathrm{Hom}}_k(H\otimes H,H)$. Thus $\varphi =\psi$ and $S(ab)=S(b)S(a)$. Evaluating $S\ast {\mathrm{i}\mathrm{d}}_H=u\epsilon$ at $1$ and using $\mathrm{\Delta }(1)=1\otimes 1$, $\epsilon (1)=1$ gives $S(1)=1$.

Uniqueness of the two-sided inverse of $\mathrm{\Delta }$ in this convolution algebra, by step 1.1, gives $\mathrm{\Phi }=\mathrm{\Psi }$, which is the asserted anti-comultiplication identity.

Since $\epsilon$ is an algebra map, applying it to $\sum S(h_{(1)})h_{(2)}=\epsilon (h)1$ gives $(\epsilon \circ S)\ast \epsilon =\epsilon$ in ${\mathrm{Hom}}_k(H,k)$. The convolution unit there is $\epsilon$, so $\epsilon \circ S=\epsilon$. Together with steps 1.1 and 2.1, this proves all assertions.