Lie algebra homomorphism

Remarks

A bijective homomorphism is an isomorphism; its inverse is again a homomorphism.

Example: let $k$ carry the zero bracket. The trace $\tr \colon \mathfrak{gl}_n(k) \to k$ is linear, and $\tr [X, Y] = \tr(XY) - \tr(YX) = 0 = [\tr X, \tr Y]$. So the trace is a homomorphism, with kernel $\mathfrak{sl}_n(k)$.

Non-example: if $n \ge 2$ and $\operatorname{char} k \ne 2$, the transpose $X \mapsto X^T$ on $\mathfrak{gl}_n(k)$ is linear but does not preserve brackets. One has $[X, Y]^T = Y^T X^T - X^T Y^T = -[X^T, Y^T]$, which differs from $[X^T, Y^T]$ whenever $[X^T, Y^T] \ne 0$, for instance for $X = E_{12}$ and $Y = E_{21}$. The map $X \mapsto -X^T$, on the other hand, is a homomorphism.

The kernel of a homomorphism is an ideal of $\mathfrak g$, and its image is a subalgebra of $\mathfrak h$.