Convolution of linear maps is associative and unital

For $f,g\in {\mathrm{Hom}}_k(C,A)$, [F2] gives the linear map $f\otimes g:C\otimes C\to A\otimes A$; composing it with ${\mathrm{\Delta }}_C$ and $m_A$ defines a linear map $f\ast g:C\to A$. The finite-sum tensor convention [F1], bilinearity of tensoring maps, and bilinearity of $m_A$ make $\ast$ bilinear in $f,g$.

For $f,g,h\in {\mathrm{Hom}}_k(C,A)$ and $c\in C$, write the finite iterated coproduct as ${\mathrm{\Delta }}_C^{(2)}(c)=\sum c_{(1)}\otimes c_{(2)}\otimes c_{(3)}$. Coassociativity from [F5] says this is the same tensor whether formed by ${\mathrm{\Delta }}_C\otimes \mathrm{i}\mathrm{d}$ or $\mathrm{i}\mathrm{d}\otimes {\mathrm{\Delta }}_C$, with the canonical reassociation justified by [F4]; associativity of $m_A$ then gives $((f\ast g)\ast h)(c)=\sum (f(c_{(1)})g(c_{(2)}))h(c_{(3)})=\sum f(c_{(1)})(g(c_{(2)})h(c_{(3)}))=(f\ast (g\ast h))(c)$. Thus $\ast$ is associative.

The left product with $u_A{\epsilon }_C$ satisfies $((u_A{\epsilon }_C)\ast f)(c)=\sum {\epsilon }_C(c_{(1)})f(c_{(2)})=f(c)$ by the left counit equation in [F5] and the unit law in $A$; the right product satisfies $(f\ast (u_A{\epsilon }_C))(c)=\sum f(c_{(1)}){\epsilon }_C(c_{(2)})=f(c)$ by the right counit equation. Hence $u_A{\epsilon }_C$ is a two-sided unit.

Steps 1.1–1.3 define a bilinear associative multiplication and verify its two-sided unit on every $c\in C$. No step requires a basis, a dual, or a dimension bound, so the result holds in arbitrary dimension.