= Solution
For unital <algebras over a field> $A,B$, a <measuring coalgebra> is a <coalgebra> $C$ together with a linear map $p:C\to\operatorname{Hom}_k(A,B)$ such that, in <Sweedler notation>,
$$
\boxed{p(c)(ab)=\sum p(c_{(1)})(a)p(c_{(2)})(b),\qquad
p(c)(1_A)=\varepsilon(c)1_B.}
$$
Thus a <group-like element> acts by a unital <algebra homomorphism over a field>, while skew-primitive elements give twisted product rules.
The <universal measuring coalgebra> $P(A,B)$ has a universal measuring map $\pi:P(A,B)\to\operatorname{Hom}_k(A,B)$ with this property: for every measuring map $p$ from a <coalgebra> $C$, there is a unique <coalgebra> homomorphism $f:C\to P(A,B)$ satisfying $\pi f=p$. The uniqueness is among <coalgebra> homomorphisms; $\pi$ itself need not be an injective map of <vector spaces>.
Back to article page
