Adjoint map is Lie homomorphism The Jacobi identity of a Lie algebra says that $ad: \mathfrak g \to End(\mathfrak g)$ is a derivation.
I am a bit emberassed but what is the easieast way to see that for every $X \in \mathfrak g$, $ad_X: \mathfrak g \to \mathfrak g, Y \mapsto [X,Y]$ is a Lie homomorphism, i.e., $[ad_X Y, ad_X Z] = ad_X([Y,Z])$ ?
Edit: The identity in the above line is wrong. It is $ad_X$ a derivation (endomorphism satisfying the product rule), not a Lie homomorphism (respecting the Lie bracket). I was a bit confused here, sorry.
 A: Continuing with @TobiasKildetoft's comment: I really think one should think that the Jacobi identity is exactly the assertion that "ad" is a Lie algebra homomorphism.
So, for example, it is not a good idea to try to present the literal Jacobi-identity formula in a "symmetric" form (as some sources do), which would necessarily obscure the assertion that "ad" is a Lie algebra hom.
It is also unfortunate that few sources seem to let on that this is what the Jacobi identity is asserting!
So, don't think of the Jacobi identity as a formula that is required to hold, but as the property that "ad" is a Lie algebra hom, ... and then you can recover the formula whenever you want.
A (slightly) more interesting related question is about why the Lie algebras of classical matrix groups do satisfy the Jacobi identity! :) That is, we really should check the property rather than accidentally taking it for granted. Of course, then the structural sense of "ad" as derivative of "Ad", which in these linear cases is "conjugation", explain it, when one spells things out.
A: $\mathrm{ad}_X$ is not a Lie Homomorphism, but $\mathrm{ad}$ is. We can define a map $\mathrm{Ad}:G\to\mathrm{Aut}(G):=\lambda x.\lambda y.\ xyx^{-1}$, whose differential is then $\mathrm{ad}:TG\to T\mathrm{Aut}(G):=\lambda X.\lambda Y.\ [X,Y]$. The homomorphism property on this map is then precisely the Jacobi identity.
