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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.

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    $\begingroup$ Why do you want to "work in the category of $\mathfrak{g}$-modules directly"? What do you even mean by that? Given that the answer to your question is $F(V)=\mathcal{U}\mathfrak{g}\otimes_k V$, it's unclear how and why you hope to avoid referring to $\mathcal{U}\mathfrak{g}$... $\endgroup$ – Eric Wofsey Sep 27 '17 at 3:07
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    $\begingroup$ Sure you can't avoid it: the free module on 1 generator is $U\mathfrak{g}$. $\endgroup$ – YCor Sep 27 '17 at 7:15
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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.

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