For the following proposition I found a proof in some notes that I don't understand. Below definition 1 defines the terminology I'm using, and proof attempt 1 gives my attempt at the proposition. Proof attempt 2 is from the notes (which are not available online), and my question is the following:
How does Proof 2 make sense? Could it be that they use another definition of quotient algebras?
Proposition: Let $g$ be a nilpotent Lie algebra, and let $h\lhd g$ be an ideal of $g$. Then the quotient Lie algebra $g/h$ is also nilpotent.
Definition 1: Let $(g, [\cdot,\cdot])$ be a Lie algebra, and $h\lhd g$ be an ideal of $g$. Then we define the quotient Lie algebra $(g/h, [\cdot, \cdot]_q)$ to be the quotient vector space with Lie bracket given by $$[\alpha+h,\beta+h]_q=[\alpha,\beta]+h.$$
Proof attempt 1 Suppose $g$ is nilpotent, $h\lhd g$. Then $C^N(g) = \{0\}$ for some $N\in\mathbb{N}$. Observe that $C^0(g/h) \subset \{\alpha+h:\ \alpha\in C^0(g)\}$, and by induction $C^n(g/h) \subset \{\alpha + h:\ \alpha \in C^n(g)\}$. It follows that $C^N(g/h) = \{0+h\}$. $\square$
Proof attempt 2: Suppose $g$ is nilpotent, $h\lhd g$. An easy induction shows that if $h$ is an ideal, then $C^n(g/h) = (C^n(g) + h)/h$, and so again if $C^N(g) = 0$ we must also have $C^N(g/h) = 0$
My difficulty is with interpreting the expression $C^n(g/h) = (C^n(g) + h)/h$, it seems erroneous to me... What am I missing?
Here is an alternative way to formulate the second attempt. Let $\mathfrak{g}$ be a nilpotent Lie algebra, and $\pi\colon \mathfrak{g}\rightarrow L$ a surjective Lie algebra homomorphism. Then $\pi(\mathfrak{g}^n)=L^n$, so that $L$ is nilpotent as well. Now chose $L=\mathfrak{g}/\mathfrak{h}$ and $\pi\colon \mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}$ the canonical surjection.