A finite calculation separating convolution from composition

The maps are defined on all three basis elements of $H$, so they extend uniquely linearly to $H$. For each basis element $r\in \{1,g,g^2\}$, [F2] gives $\mathrm{\Delta }(r)=r\otimes r$, hence $(f\ast q)(r)=f(r)q(r)$. The products for $r=1,g,g^2$ are respectively $0\cdot g=0$, $1\cdot 0=0$, and $0\cdot 0=0$, so $f\ast q$ is the zero linear map.

By [F2], $S(g)=g^{-1}=g^2\mathrm{\ne }g={\mathrm{i}\mathrm{d}}_H(g)$, so $S\mathrm{\ne }{\mathrm{i}\mathrm{d}}_H$. The only compositional inverse of ${\mathrm{i}\mathrm{d}}_H$ is itself: if $T\circ {\mathrm{i}\mathrm{d}}_H={\mathrm{i}\mathrm{d}}_H$, then $T={\mathrm{i}\mathrm{d}}_H$. Yet [F3] and the Hopf structure make $S$ the two-sided convolution inverse of ${\mathrm{i}\mathrm{d}}_H$. The inverse operation in the Hopf definition is therefore convolution, not composition.

By definition of composition, $(f\circ q)(1)=f(q(1))=f(g)=1$, which is nonzero in the basis of $H$. Thus $f\circ q$ is not the zero map and differs from $f\ast q$ in step 1.1.

Steps 1.1, 1.2, and 2.1 give an explicit pair of maps with different convolution and composition products and a Hopf antipode that is not a composition inverse of the identity.