Suppose I have a Lie algebra $\mathfrak g$ which has an ideal $\mathfrak a$. Then I consider the quotient set $\mathfrak g / \mathfrak a$ which is the set of all equivalence relations of $\mathfrak g$ on $\mathfrak a$, where the appropriate equivalence relation is $v \sim u \iff v-u \in \mathfrak a$. The quotient set is also a vector space if we equip it with the following scalar multiplication and vector addition:
$$ \begin{align*} \lambda [u] &= [\lambda u] \\ [u] + [v] &= [u+v] \end{align*} $$
where $[u]$ is the equivalence class of $u$.
Now, we can also extended $\mathfrak g / \mathfrak a$ to be a Lie algebra as well if we equip it with a Lie bracket $\{\}$ defined by:
$$ \{[u],[v]\} = [\langle u,v \rangle] $$
where $\langle \rangle$ is the Lie brackets for the Lie algebra $\mathfrak g$.
My question is: is this unique? Is $\{\}$ the only Lie bracket I could possibly define for $\mathfrak g / \mathfrak a$, or is there another one?
Nate's comment is correct, but here is a bit more to chew on.
Notice that if $\mathfrak{h}$ is any Lie algebra such that there is an isomorphism of vector spaces $f:\mathfrak{h}\to\mathfrak{g}/\mathfrak{a}$, then we can define a Lie algebra structure on $\mathfrak{g}/\mathfrak{a}$ by
$$\{[u],[v]\}=f(\langle a,b\rangle)$$
where $f(a)=[u]$ and $f(b)=[v]$. Notice this is well-defined because $f$ is an isomorphism. In this definition, we are simply translating structure along the isomorphism $f$, turning $f$ into a map of Lie algebras. In particular, we can always equip the vector space $\mathfrak{g}/\mathfrak{a}$ with the trivial bracket, $\{[u],[v]\}=0$ for all $[u],[v]\in\mathfrak{g}/\mathfrak{a}$. However, in this general construction, we've forgotten the particularly nice structure of the object $\mathfrak{a}$, which is a Lie ideal, and not just a vector subspace.
To take advantage of the Lie ideal structure, it turns out that the natural surjection of vector spaces $f':\mathfrak{g}\to\mathfrak{g}/\mathfrak{a}$, sending $u$ to $[u]$, induces a Lie algebra structure on $\mathfrak{g}/\mathfrak{a}$ just as defined above:
$$\{[u],[v]\}=f'(\langle u,v\rangle)=[\langle u,v\rangle]$$
In our first example, we used the fact that $f$ was an isomorphism to show that our bracket was well-defined, however, $f'$ need not be an isomorphism. Instead, to see that this bracket is well-defined, you'll need exactly what the Lie ideal structure provides. As before, we'll see then that $f'$ becomes a map of Lie algebras, not just of vector spaces. Thus, this is the most natural bracket to place on the object $\mathfrak{g}/\mathfrak{a}$, as it seems the conditions required of $\mathfrak{a}$ were predetermined specifically for this construction.