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As I understand the question the poser wanted a construction of the envelopping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.

Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the envelopping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthhermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$

Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermor, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ correspond here also to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$ T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g) $$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions show that it is indeed a Lie algebra morphism.

As I understand the question the poser wanted a construction of the enveloping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.

Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the enveloping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$

Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermore, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ again corresponds to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$ T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g) $$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions shows that it is indeed a Lie algebra morphism.

[[There is some overlap with Theo's answer but I have not tried to take that into consideration. [Except now for the addendum.] ]]

Well, now that we have been placed ourselves in characteristic zero we can, in the usual case, use the canonical isomorphism $S^\ast\mathfrak{g}\rightarrow U(\mathfrak{g})$ given by the composite $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n} \rightarrow U(\mathfrak{g})$, with the last map given by multiplication. This imbues $S^\ast\mathfrak{g}$ with a funny multiplication and if we can express it using only the Lie bracket and tensor concatenation we can extend it to a general symmetric monoidal category context. The twisted multiplication is given by a bunch of maps $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \rightarrow S^k\mathfrak{g}$ (where it is easy to see that the maps are zero unless $k\leq i+j$). It is enough to consider the $n$-product map $\mathfrak{g}^{\otimes n}\rightarrow S^k\mathfrak{g}$ is the general map is the composite of $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \subseteq \mathfrak{g}^{\otimes i+j} \rightarrow S^k\mathfrak{g}$. To get the idea that there are universal formulas we look at small values: We have $u\otimes v=u\odot v+1/2[u,v]$, where $u\odot v$ is the symmetrisation $1/2(u\otimes v+v\otimes u)$. Similarly, considering $u\odot v\odot w$ we may systematically transpose the summands distinct from $u\otimes v\otimes w$ at the expense of putting in a commutator to get (for instance) $$ u\odot v\odot w = u\otimes v\otimes w + \frac{1}{2}u\otimes [w,v]+v\otimes[w,u]+w\otimes[v,u]+\frac{1}{3}([u,v]\otimes v+[u,w]\otimes w). $$ We can then recursively reduce the rest of the terms. (Note that the exact formulas depend exactly how we perform the reduction.) This continues in higher degrees and give specific (making as I said some specific choices for the reductions but the Jacobi identity and the anti-symmetry imply that the result is independent of those choices). we can now transfer these formulas to a general symmetric monoidal category.

It seems not at all clear that you get an associative algebra in this way but I think one should be able to prove that it is a formal consequence of the anti-symmetry and the Jacobi identity by considering the free Lie algebra on varying vector spaces.

Addendum: As Theo points out, the enveloping algebra should preferably have a universal property. Giving a Lie algebra morphism $\mathfrak{g} \rightarrow A$ into an associative algebra object we can get maps $S^i\mathfrak{g} \rightarrow A$ by composing the inclusion $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n}$, the induced map $\mathfrak{g}^{\otimes n} \rightarrow A^{\otimes n}$ and the product map of $A$. That this gives an algebra map $U(\mathfrak{g}) \rightarrow A$ comes down to some universal identities which again one should be able to verify by reduction to the ordinary case.

I have now understood the situation better so my previous post has been replaced by this. (The only thing that was in the original but will not be here are some explicit formulas but Theo has given a reference for that.)

As I understand the question the poser wanted a construction of the envelopping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.

Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the envelopping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthhermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$

Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermor, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ correspond here also to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$ T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g) $$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions show that it is indeed a Lie algebra morphism.

Finally assuming that $\mathfrak g \to A$ is a Lie algebra homomorphism where $A$ is an associative algebra with $L \to T$ inducing its Lie algebra structure. Note that we have an isomorphism (now going back to vector spaces) $S(L(V))\to T(V)$ and hence an isomorphism is $\Sigma$-modules $S\circ L=T$. This gives us a map $S(\mathfrak g)\to S(L(\mathfrak g))=T(\mathfrak g) \to T(A) \to A$ and it is easy to see that this is an algebra morphism.

[[There is some overlap with Theo's answer but I have not tried to take that into consideration. ]]

Well, now that we have been placed ourselves in characteristic zero we can, in the usual case, use the canonical isomorphism $S^\ast\mathfrak{g}\rightarrow U(\mathfrak{g})$ given by the composite $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n} \rightarrow U(\mathfrak{g})$, with the last map given by multiplication. This imbues $S^\ast\mathfrak{g}$ with a funny multiplication and if we can express it using only the Lie bracket and tensor concatenation we can extend it to a general symmetric monoidal category context. The twisted multiplication is given by a bunch of maps $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \rightarrow S^k\mathfrak{g}$ (where it is easy to see that the maps are zero unless $k\leq i+j$). It is enough to consider the $n$-product map $\mathfrak{g}^{\otimes n}\rightarrow S^k\mathfrak{g}$ is the general map is the composite of $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \subseteq \mathfrak{g}^{\otimes i+j} \rightarrow S^k\mathfrak{g}$. To get the idea that there are universal formulas we look at small values: We have $u\otimes v=u\odot v+1/2[u,v]$, where $u\odot v$ is the symmetrisation $1/2(u\otimes v+v\otimes u)$. Similarly, considering $u\odot v\odot w$ we may systematically transpose the summands distinct from $u\otimes v\otimes w$ at the expense of putting in a commutator to get (for instance) $$ u\odot v\odot w = u\otimes v\otimes w + \frac{1}{2}u\otimes [w,v]+v\otimes[w,u]+w\otimes[v,u]+\frac{1}{3}([u,v]\otimes v+[u,w]\otimes w). $$ We can then recursively reduce the rest of the terms. (Note that the exact formulas depend exactly how we perform the reduction.) This continues in higher degrees and give specific (making as I said some specific choices for the reductions but the Jacobi identity and the anti-symmetry imply that the result is independent of those choices). we can now transfer these formulas to a general symmetric monoidal category.

It seems not at all clear that you get an associative algebra in this way but I think one should be able to prove that it is a formal consequence of the anti-symmetry and the Jacobi identity by considering the free Lie algebra on varying vector spaces.

[[There is some overlap with Theo's answer but I have not tried to take that into consideration. [Except now for the addendum.] ]]

Well, now that we have been placed ourselves in characteristic zero we can, in the usual case, use the canonical isomorphism $S^\ast\mathfrak{g}\rightarrow U(\mathfrak{g})$ given by the composite $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n} \rightarrow U(\mathfrak{g})$, with the last map given by multiplication. This imbues $S^\ast\mathfrak{g}$ with a funny multiplication and if we can express it using only the Lie bracket and tensor concatenation we can extend it to a general symmetric monoidal category context. The twisted multiplication is given by a bunch of maps $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \rightarrow S^k\mathfrak{g}$ (where it is easy to see that the maps are zero unless $k\leq i+j$). It is enough to consider the $n$-product map $\mathfrak{g}^{\otimes n}\rightarrow S^k\mathfrak{g}$ is the general map is the composite of $S^i\mathfrak{g}\bigotimes S^j\mathfrak{g} \subseteq \mathfrak{g}^{\otimes i+j} \rightarrow S^k\mathfrak{g}$. To get the idea that there are universal formulas we look at small values: We have $u\otimes v=u\odot v+1/2[u,v]$, where $u\odot v$ is the symmetrisation $1/2(u\otimes v+v\otimes u)$. Similarly, considering $u\odot v\odot w$ we may systematically transpose the summands distinct from $u\otimes v\otimes w$ at the expense of putting in a commutator to get (for instance) $$ u\odot v\odot w = u\otimes v\otimes w + \frac{1}{2}u\otimes [w,v]+v\otimes[w,u]+w\otimes[v,u]+\frac{1}{3}([u,v]\otimes v+[u,w]\otimes w). $$ We can then recursively reduce the rest of the terms. (Note that the exact formulas depend exactly how we perform the reduction.) This continues in higher degrees and give specific (making as I said some specific choices for the reductions but the Jacobi identity and the anti-symmetry imply that the result is independent of those choices). we can now transfer these formulas to a general symmetric monoidal category.

It seems not at all clear that you get an associative algebra in this way but I think one should be able to prove that it is a formal consequence of the anti-symmetry and the Jacobi identity by considering the free Lie algebra on varying vector spaces.

Addendum: As Theo points out, the enveloping algebra should preferably have a universal property. Giving a Lie algebra morphism $\mathfrak{g} \rightarrow A$ into an associative algebra object we can get maps $S^i\mathfrak{g} \rightarrow A$ by composing the inclusion $S^n\mathfrak{g}\subseteq \mathfrak{g}^{\otimes n}$, the induced map $\mathfrak{g}^{\otimes n} \rightarrow A^{\otimes n}$ and the product map of $A$. That this gives an algebra map $U(\mathfrak{g}) \rightarrow A$ comes down to some universal identities which again one should be able to verify by reduction to the ordinary case.