Lie representations are U(g)-modules

A representation $\rho :\mathfrak{g}\to {\mathrm{End}}_k(V{)}_{\mathrm{L}\mathrm{i}\mathrm{e}}$ extends uniquely by [L2] to a unital algebra map $\bar{\rho }:U(\mathfrak{g})\to {\mathrm{End}}_k(V)$, hence to a unital left module structure by [L1].

Conversely, a scalar-compatible unital module gives a $k$-algebra map $\alpha :U(\mathfrak{g})\to {\mathrm{End}}_k(V)$. Its restriction $\alpha {\iota }_{\mathfrak{g}}$ preserves Lie brackets because both ${\iota }_{\mathfrak{g}}$ and every algebra map preserve commutators, so it is a representation. Starting with $\rho$ recovers $\rho$ by $\bar{\rho }{\iota }_{\mathfrak{g}}=\rho$; starting with $\alpha$ recovers $\alpha$ by uniqueness in [L2].

Let $T:V\to W$ be linear. If $T$ intertwines the $\mathfrak{g}$-actions, then it intertwines each $\rho (x)$, each finite product of these operators, and finite linear combinations of the products. By [L4], these include the actions of every element of $U(\mathfrak{g})$; the empty product acts as the identity on both modules. Thus $T$ is a module homomorphism. Conversely, restricting a module homomorphism to ${\iota }_{\mathfrak{g}}(\mathfrak{g})$ gives an intertwiner.

The object and morphism correspondences in steps 1.1–1.3 are mutually inverse, proving the claimed equivalence without any PBW or injectivity assumption.