Counit uniqueness and finite Sweedler calculus

For $c\in C$, write one finite expression $\mathrm{\Delta }(c)={\sum }_i{a}_i\otimes b_i$. The scalar ${\sum }_i\epsilon (a_i)\eta (b_i)$ is the value on $\mathrm{\Delta }(c)$ of the linear contraction ${\mu }_k\circ (\epsilon \otimes \eta )$, where ${\mu }_k:k\otimes k\to k$ is scalar multiplication; it is therefore independent of the chosen expression by [F2] and [F3]. Applying $\eta$ to the left counit equation for $\epsilon$ shows this scalar is $\eta (c)$, while applying $\epsilon$ to the right counit equation for $\eta$ shows it is $\epsilon (c)$. Hence $\epsilon (c)=\eta (c)$ for every $c$.

For any linear maps $f:C\to X$ and $g:C\to Y$, the expression ${\sum }_{i}f(a_i)\otimes g(b_i)$ equals $(f\otimes g)\mathrm{\Delta }(c)$ by [F3]. Thus changing the finite tensor expression for $\mathrm{\Delta }(c)$ cannot change this contraction. Iterated Sweedler notation similarly records the tensor obtained from a specified composite of comultiplications; [F4] reparenthesizes its target. If $\mathrm{\Delta }$ is coassociative as in Coalgebra and coalgebra map over a field, then $\alpha (\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }$, so these two triple coproducts agree after reparenthesization. Counit uniqueness above requires no coassociativity.