Universal property of the enveloping algebra

By [L1], $f$ extends uniquely to a unital algebra homomorphism $F:T(\mathfrak{g})\to A$. Since $f$ preserves Lie brackets, $F(x\otimes y-y\otimes x-[x,y])=f(x)f(y)-f(y)f(x)-f([x,y])=0$ for every defining relator.

The kernel of $F$ is a two-sided ideal containing all defining relators, hence contains their generated ideal $I$. By [L2], $F$ factors uniquely as $T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})\stackrel{\bar{f}}{\to }A$, and the factor satisfies $\bar{f}{\iota }_{\mathfrak{g}}=f$.

If $G:U(\mathfrak{g})\to A$ is another unital algebra map with $G{\iota }_{\mathfrak{g}}=f$, its composite with $T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})$ is an algebra extension of $f$. It equals $F$ by [L1], and quotient-map surjectivity gives $G=\bar{f}$.

Thus the required extension exists uniquely; the argument uses only the quotient presentation and not injectivity of ${\iota }_{\mathfrak{g}}$.