[\imath_{\mathfrak{g}}(X),\imath_A(a)] = \imath_A({\alpha}(X)a);$$ and $({\mathcal{U}}({\mathcal{A}},{\mathfrak{g}}),\imath_{\mathcal{A}},\imath_{\mathfrak{g}})$ is universal among these triples; that is, for a similar triple $(B,\phi_{\mathcal{A}},\phi_{\mathfrak{g}})$, there is a unique algebra morphism $\Phi_B:{\mathcal{U}}({\mathcal{A}},{\mathfrak{g}})\to B$ such that $\Phi_B\imath_{\mathcal{A}}=\phi_{\mathcal{A}}$ and $\Phi_B\imath_{\mathfrak{g}}=\phi_{\mathfrak{g}}$.

To construct ${\mathcal{U}}({\mathcal{A}},{\mathfrak{g}})$ we employ ${\mathcal{U}}({\mathfrak{g}})$ in the following way. The condition that the anchor maps into derivations means precisely that ${\mathcal{A}}$ is a Hopf ${\mathcal{U}}({\mathfrak{g}})$-module [@Quaoar]. One may keep using the same notation ${\alpha}$ for the new action, as for the generators; ${\alpha}(1)=1$. Consider now the smash product or crossed product algebra ${\mathcal{A}}\rtimes{\mathcal{U}}({\mathfrak{g}})$. This is the vector space ${\mathcal{A}}{\otimes}{\mathcal{U}}({\mathfrak{g}})$ with the product defined on simple tensors by $$(a {\otimes}u)(b {\otimes}v) :=  a{\alpha}(u_{(1)})b {\otimes}u_{(2)}v,$$ where we use the standard Sweedler notation ${\Delta}u=u_{(1)}{\otimes}u_{(2)}$. There are obvious morphisms $\imath'_{\mathcal{A}}:{\mathcal{A}}\to{\mathcal{A}}\rtimes
{\mathcal{U}}({\mathfrak{g}})$ and $\imath'_{\mathfrak{g}}:{\mathfrak{g}}\to{\mathcal{A}}\rtimes{\mathcal{U}}({\mathfrak{g}})$. Now, let $J$ be the right ideal generated in ${\mathcal{A}}\rtimes U({\mathfrak{g}})$ by the elements $ab{\otimes}X -
a{\otimes}bX$. One has $$\begin{aligned}
(c {\otimes}X)(ab {\otimes}Y - a {\otimes}bY) &= cab {\otimes}XY + c{\alpha}(X)(ab) {\otimes}Y - ca
{\otimes}XbY - c{\alpha}(X)a {\otimes}bY
\\
&= cab {\otimes}XY + c{\alpha}(X)a\,b {\otimes}Y + ca{\alpha}(X)b {\otimes}Y
\\
&\quad -ca {\otimes}bXY - ca {\otimes}{\alpha}(X)b\,Y - c{\alpha}(X)a {\otimes}bY
\\
&= cab{\otimes}XY - ca {\otimes}bXY + c{\alpha}(X)a\,b {\otimes}Y - c{\alpha}(X)a {\otimes}bY
\\
&\quad + ca{\alpha}(X)b {\otimes}Y - ca {\otimes}{\alpha}(X)b\,Y,\end{aligned}$$ where in 