Let $\mathfrak{g}$ be a Lie algebra over a field $k$, we can consider the category of $\mathfrak{g}$-modules. I want to know that what are the free objects in this category. Or do we have a free construction functor $F$ assigning each vector space over $k$ to a $\mathfrak{g}$-module such that we have the following natural bijection $$\mathrm{Hom}_{k}(V,\,M)\cong \mathrm{Hom}_{\mathfrak{g}}(F(V),\,M)$$ where $M$ is any $\mathfrak{g}$-module.
I know the equivalence between the category of $\mathfrak{g}$-modules and the category of left $\mathcal{U}\mathfrak{g}$-modules, but I want to work in the category of $\mathfrak{g}$-modules directly.
Any help or hints would be very appreciate.
You're going to have to work with $U(\mathfrak g)$ whether you like it or not because the categories of $\mathfrak g$-modules and $U(\mathfrak g)$-modules aren't just equivalent, they are isomorphic on the nose. You know that in the category of $U(\mathfrak g)$-modules the functor $F$ you're looking for is $F(V) = U(\mathfrak g) \otimes_k V$ so there is no other choice for the category of $\mathfrak g$-modules, the free $\mathfrak g$-modules are exactly the modules of the form $U(\mathfrak g) \otimes_k V$.
On the plus side, the isomorphism between those two categories is the identity on the underlying vector spaces, it just switches out the additional structures. So you can take your free modules $U(\mathfrak g) \otimes_k V$ and forget that $U(\mathfrak g)$ acts on them and just work with them as $\mathfrak g$-modules if that's what you'd like.