Definition 5

Let $a$ and $b$ be two Lie algebras over $K$. An extension of $b$ by $a$ is a sequence:

where $g$ is a Lie algebra over $K$, $\mu$ a surjective homomorphism of $g$ onto $b$ and $\lambda$ an injective homomorphism of $a$ onto the kernel of $\mu$.

The kernel $n$ of $\mu$ is called the kernel of the extension. The homomorphism $\lambda$ is an isomorphism of $a$ onto $n$ and the homomorphism $\mu$ defines an isomorphism of $g / n$ onto $b$ when passing to the quotient. By an abuse of language, $g$ is also called an extension of $b$ by $a$.

Two extensions:

are said to be equivalent if there exists a homomorphism $f$ of $g$ into $g'$ such that the following diagram:

is commutative (that is such that $f \circ \lambda = \lambda', \mu' \circ f = \mu$). We show that such a homomorphism is necessarily bijective. First $f$ is injective. For if $x \in g$ is such that $f(x) = 0$, then $\mu(x) = \mu'(f(x)) = 0$ and hence $x = \lambda(y)$ for some $y \in a$; then $\lambda'(y) = f(\lambda(y)) = f(x) = 0$, hence $y = 0$ and hence $x = 0$. On the other hand, $f$ is surjective. For $\mu' \circ f = \mu$ is surjective and hence $f(g) + \lambda'(a) = g'$; on the other hand $f(g) \supset f(\lambda(a)) = \lambda'(a)$.

It follows from this that the relation just defined between two extensions of $b$ by $a$ is an equivalence relation.