The quotient Lie-algebra bracket is well-defined

If $x^{\mathrm{\prime }}=x+a$ and $y^{\mathrm{\prime }}=y+b$ with $a,b\in \mathfrak{i}$, bilinearity gives $[x^{\mathrm{\prime }},y^{\mathrm{\prime }}]-[x,y]=[a,y]+[x,b]+[a,b]$. Every term belongs to $\mathfrak{i}$ by [L1], so $[x^{\mathrm{\prime }},y^{\mathrm{\prime }}]+\mathfrak{i}=[x,y]+\mathfrak{i}$.

Bilinearity of the quotient bracket follows by choosing representatives, using bilinearity in $\mathfrak{g}$, and using [L2] for coset addition and scalar multiplication; step 1.1 makes the result independent of those choices.

For every $x$, one has $[x+\mathfrak{i},x+\mathfrak{i}]=[x,x]+\mathfrak{i}=\mathfrak{i}$, so the descended bracket is alternating.

For three cosets, their cyclic Jacobi sum is the coset of $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]$, which is zero in $\mathfrak{g}$ and hence is the zero coset. Thus Jacobi descends.

Therefore the displayed rule has all Lie identities. For $\mathfrak{i}=0$ it is the original bracket, and for $\mathfrak{i}=\mathfrak{g}$ it is the unique bracket on the zero space; no choice of representatives is made in either case or in the argument above.