Let $\mathcal{L}$ denote the functor $AssAlg \to LieAlg$ which sends each associative unital algebra to the same underlying vector space equipped with a bracket defined as $[x,y] = xy - yx$ and $\mathcal{U}$ denote the functor $LieAlg \to AssAlg$ which sends each Lie algebra $L$ to $\mathcal{T}(L)/I$, the tensor algebra of $L$ quotiented by the ideal $I$ generated by elements of the form $x \otimes y - y \otimes x - [x,y]$.
I'm interested in defining an unital associative algebra homomorphism $\mathcal{U}\mathcal{L}(A) \to A$ satisfying a certain universal property. However, I can't see how should this homomorphism be defined. My idea was to define some "natural" homomorphism $\mathcal{T}(\mathcal{L}(A)) \to A$ which kernel would contain $I$, hence obtaining a homomorphism $\mathcal{U}\mathcal{L}(A) \to A$ making the appropriate diagram commute. However, i can't prove the ideal inclusion when the map $\mathcal{T}(\mathcal{L}(A)) \to A$ is simply a direct sum of projections. I don't know if this is the way to exhibit the co-unit either.
Let $\mathfrak{g}$ be a Lie algebra and let $\mathcal{U}(\mathfrak{g}) = \mathcal{T}(\mathfrak{g}) / I$ be the universal enveloping algebra of $\mathfrak{g}$. We denote for every element $t$ of $\mathcal{T}(\mathfrak{g})$ by $[t]$ the corresponding residue class in $\mathcal{U}(\mathfrak{g})$.
The universal property of the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$ states that there exists for every associative, unitial algebra $A$ and every homomorphism of Lie algebras $\varphi$ from $\mathfrak{g}$ to $\mathcal{L}(A)$ a unique homomorphism of algebras $\Phi$ from $\mathcal{U}(\mathfrak{g})$ to $A$ which extends $\varphi$. That $\Phi$ is an extension of $\varphi$ means that $$ \Phi( [x] ) = \varphi(x) $$ for all $x \in \mathfrak{g}$. It then further follows that \begin{align*} \Phi( [x_1 \otimes \dotsb \otimes x_n] ) &= \Phi( [x_1] \dotsm [x_n] ) \\ &= \Phi( [x_1] ) \dotsm \Phi( [x_n] ) \\ &= \varphi(x_1) \dotsm \varphi(x_n) \end{align*} for all $x_1, \dotsc, x_n \in \mathfrak{g}$.
To construct this homomorphism $\Phi$ one starts by the universal property of the tensor algebra $\mathcal{T}(\mathfrak{g})$ with a homomorphism of algebras $$ \Psi \colon \mathcal{T}(\mathfrak{g}) \to A $$ that extends $\varphi$. This homomorphism $\Psi$ is given by $$ \Psi( x_1 \otimes \dotsb \otimes x_n ) = \varphi(x_1) \dotsm \varphi(x_n) $$ for all $x_1, \dotsc, x_n \in \mathfrak{g}$. The ideal $I$ is contained in the kernel of $\Psi$ because \begin{align*} \Psi( x \otimes y - y \otimes x ) &= \varphi(x) \varphi(y) - \varphi(y) \varphi(x) \\ &= [ \varphi(x), \varphi(y) ] \\ &= \varphi( [x,y] ) \\ &= \Psi( [x,y] ) \,, \end{align*} where we use for the second to last equality that $\varphi$ is a homomorphism of Lie algebras from $\mathfrak{g}$ to $\mathcal{L}(A)$. It follows that $\Psi$ induces a homomorphism of algebras $\Phi$ from $\mathcal{U}( \mathfrak{g} )$ to $A$, as required.
Suppose now that $A$ is an associative, unital algebra. The identity map of $\mathcal{L}(A)$, given by $$ \varphi \colon \mathcal{L}(A) \to \mathcal{L}(A) \,, \quad a \mapsto a $$ is a homomorphism of Lie algebras. The induced homomorphism of algebras $$ \Phi \colon \mathcal{U}(\mathcal{L}(A)) \to A $$ is precisely the counit of the adjunction $\mathcal{U} \dashv \mathcal{L}$. It is by the above discussion given by $$ \Phi( [ a_1 \otimes \dotsb \otimes a_n ] ) = \varphi(a_1) \dotsm \varphi(a_n) = a_1 \dotsm a_n $$ for all $a_1, \dotsc, a_n \in A$.