+ a{\mathcal{L}}_X {\otimes}{\mathcal{L}}_Y + {\mathcal{L}}_Y {\otimes}a{\mathcal{L}}_X + 1
{\otimes}a{\mathcal{L}}_X{\mathcal{L}}_Y {\quad\mbox{and}\quad}
\\
{\delta}(a{\mathcal{L}}_Y{\mathcal{L}}_X) &= a{\mathcal{L}}_Y{\mathcal{L}}_X {\otimes}1 + a{\mathcal{L}}_Y {\otimes}{\mathcal{L}}_X + {\mathcal{L}}_X {\otimes}a{\mathcal{L}}_Y +
1 {\otimes}a{\mathcal{L}}_Y{\mathcal{L}}_X.\end{aligned}$$ Now, the right hand sides are not equal. As a corollary we have that the map ${\mathcal{L}}$ is not injective: the fact that in ${\mathcal{U}}({\mathfrak{X}}(M))$ one has a good coproduct implies that $aX{\cdot}Y$ is there different &gt;from $aY{\cdot}X$, with ${\cdot}$ the product in the enveloping algebra. Thus the kernel of ${\mathcal{L}}$ contains $aX{\cdot}Y-aY{\cdot}X$. Notice that the argument fails if $M$ is one-dimensional. Notice as well that ${\delta}$ makes ${\mathbb{D}}(M)$ into a good coalgebra over ${\mathcal{F}}(M)$. However, the product is then not ${\mathcal{F}}(M)$-linear.

In spite of the above, Hopf algebra renders us a first great service in helping to manufacture ${\mathbb{D}}(M)$ out of ${\mathcal{U}}({\mathfrak{X}}(M))$. This involves a purely algebraic construction [@Bigotes1; @Harbinger] suggested by the previous discussion and better presented in the context of Lie–Rinehart algebras.

Assume for simplicity that ${\mathcal{A}}$ is unital. The *universal object* of $({\mathcal{A}},{\mathfrak{g}})$ is by definition a triple $({\mathcal{U}}({\mathcal{A}},{\mathfrak{g}}),\imath_{\mathcal{A}},
\imath_{\mathfrak{g}})$, where ${\mathcal{U}}({\mathcal{A}},{\mathfrak{g}})$ is an associative algebra, therefore a Lie algebra with the usual commutator, together with morphisms $\imath_{\mathcal{A}}:{\mathcal{A}}\to{\mathcal{U}}({\mathcal{A}},{\mathfrak{g}})$ and $\imath_{\mathfrak{g}}:{\mathfrak{g}}\to{\mathcal{U}}({\mathcal{A}},{\mathfrak{g}})$, respectively of algebras and Lie algebras, such that $$\imath_A(a)\imath_{\mathfrak{g}}(X) = \imath_{\mathfrak{g}}(aX); \qquad
