Let $\mathfrak{g}$ be a Lie algebra, and $\mathfrak{h} \subset \mathfrak{g}$ a Linear subspace which is an ideal. That is, for every $X \in \mathfrak{h}$ and every $Y \in \mathfrak{g}$, one has $[X,Y] \in \mathfrak{h}$.
(i) Explain why every ideal is a Lie subalgebra.
(ii) If $\mathfrak{h} \subset \mathfrak{g}$ is an ideal prove there exists a unique Lie algebra structure such that the canonical projection map
$$\pi : \mathfrak{g} \to \mathfrak{g} / \mathfrak{h}$$
is a Lie algebra homomorphism.
Attempt:
For (i), we just need to show $\mathfrak{h}$ is closed under the bracket operation, but if $X,Y \in \mathfrak{h}$, then $Y \in \mathfrak{g}$ and as $\mathfrak{h}$ is an ideal, one has $[X,Y] \in \mathfrak{h}$. Thus $\mathfrak{h} \subset \mathfrak{g}$ is a Lie subalgebra.
For (ii) I am a bit more lost, so is this saying, given $\pi$ is a Lie algebra homomorphism, then show the brackets operation is unique. Does this resort to showing $\pi$ is well-defined? Because I know $\pi$ is a homomorphism iff for all $X,Y \in \mathfrak{g}$, one has
$$\pi[X,Y]_\mathfrak{g}=[\pi(X),\pi(Y)]_{\mathfrak{g}/\mathfrak{h}}.$$
I.e., if you apply $\pi$, then brackets its the same as applying bracket then $\pi$. This holds iff
$$[X,Y]_{\mathfrak{g}/\mathfrak{h}}=\pi([\pi^{-1}(X),\pi^{-1}(Y)]_\mathfrak{g}).$$
Also, this may be silly, but $\pi$ sends $g$ to the equivalence class $g+H$, correct? Thanks for any hints or tips, I am new to diff geo.
What you write looks correct (you mean "$\pi$ sends $g$ to the equivalence class $g +\mathfrak{h}$"), but maybe it helps to rephrase it this way: For $\pi$ to be a homomorphism, as you say, we are forced to define
$$[X + \mathfrak h, Y + \mathfrak h]_{\mathfrak{g}/\mathfrak{h}} := [X,Y]_\mathfrak{g} +\mathfrak h$$
but for this definition to make sense (i.e. giving a well-defined bracket on the quotient), we need that for all $X_1, X_2, Y_1, Y_2 \in \mathfrak g$:
$$\left(X_1 + \mathfrak h = X_2 + \mathfrak h \text{ and } Y_1 + \mathfrak h = Y_2 + \mathfrak h\right) \implies [X_1,Y_1]+ \mathfrak h = [X_2, Y_2] + \mathfrak h$$
This is what you want to check, using that $\mathfrak h$ is an ideal.