We call a sequence of $L_\infty$-algebras (weak) maps
$$0\to L\xrightarrow{f} M\xrightarrow{g} N\to 0$$
is *exact* if it is exact on the the underlying chain complexes level.

Thought I don't know whether this is a good notion. A trivial case is exact sequence of Lie algebras. It is clear that given a surjective map of Lie algebras, the kernel is a Lie algebra as well.

My naive question: if we are only given $L_\infty$-algebras $$M\xrightarrow{g} N\to 0$$ being exact, can we find a $L_\infty$-algebra $L$ fitting the longer sequence above?

I expect the answer to this naive question is no, but I do not have an example. If the answer is no, then how do one sensibly fix it? I heard someone talked about homotopy fibre of $L_\infty$-algebras, does it work for this case?