Corollary

If u is injective (resp. surjective, bijective), so is $L(u)$.

Since the assertion is trivial for $X = \varnothing$, we assume $X \neq \varnothing$. If u is injective there exists a mapping v of Y into X such that $v \circ u$ is the identity mapping of X; by Proposition 2, $L(v) \circ L(u)$ is the identity of automorphism of $L(X)$ and hence $L(u)$ is injective. When u is surjective, there exists a mapping w of Y into X such that $u \circ w$ is the identity mapping of Y; then $L(u) \circ L(w)$ is the identity mapping of $L(Y)$, which proves that $L(u)$ is surjective.

† Recall (Algebra, Chapter IV, § 5, no. 10) the definition of polynomial mappings of a free module M into a module N: if q is an integer $\geq 0$, a mapping $f : M \to N$ is called homogeneous polynomial of degree q if there exists a multilinear mapping u of $M^q$ into N such that

A mapping of M into N is called polynomial if it is a finite sum of homogeneous polynomial mappings of suitable degrees.

Let X be a set and S a subset of X. The above corollary shows that the canonical injection of S into X can be extended to an isomorphism $\alpha$ of $L(S)$ onto the Lie subalgebra $L'(S)$ of $L(X)$ generated by S; we shall identify $L(S)$ and $L'(S)$ by means of $\alpha$.

Let $(S_\alpha)_{\alpha \in I}$ be a right directed family of subsets of X with union S. The relation $S_\alpha \subset S_\beta$ implies $L(S_\alpha) \subset L(S_\beta)$ and hence the family of Lie subalgebras $L(S_\alpha)$ of $L(X)$ is right directed. Therefore $g = \bigcup_{\alpha \in I} L(S_\alpha)$ is a Lie subalgebra of $L(X)$; then $S \subset g$, whence $L(S) \subset g$, and, as $L(S_\alpha) \subset L(S)$ for all $\alpha \in I$, $g \subset L(S)$. Hence

for every right directed family $(S_\alpha)_{\alpha \in I}$ of subsets of X.

Applying the above to the family of finite subsets of X, we see that every element of $L(X)$ is of the form $P(x_1, \ldots, x_n)$ where P is a Lie polynomial in n indeterminates and $x_1, \ldots, x_n$ are elements of X.