Proposition 2

Let X and Y be two sets. Every mapping $u : X \to Y$ can be extended uniquely to a Lie algebra homomorphism $L(u) : L(X) \to L(Y)$. For every mapping $v : Y \to Z$, $L(v \circ u) = L(v) \circ L(u)$.

The existence and uniqueness of $L(u)$ follow from Proposition 1 of no. 2. The homomorphisms $L(v \circ u)$ and $L(v) \circ L(u)$ have the same restriction to X and hence are equal (Proposition 1).