Let $\mathfrak g$ be a Lie algebra (if it matters, right now I only care about finite-dimensional Lie algebras in characteristic $0$, although I'm never opposed to hearing about more general cases). Recall that it determines a differential graded algebra ("the complex that computes Lie algebra cohomology"), with $k$th component $\wedge^k \mathfrak g^\*$ and differential determined by the bracket. It is a complex by the Jacobi identity. Thinking in terms of supergeometry, I will call this dga $\mathcal C^\infty([-1]\mathfrak g)$.


Moreover, if $\mathfrak h \to \mathfrak g$ is a homomorphism of Lie algebras, then we get a homomorphism of dgas $\mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$, so that $\mathcal C^\infty\circ [-1]$ is a contravariant functor. The dga is a complete invariant: the functor is full and faithful.



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> Suppose that $\mathfrak h,\mathfrak g$ are two Lie algebras and $f: \mathfrak h \to \mathfrak g$ a Lie algebra homomorphism. Suppose furthermore that the corresponding map $f^\*: \mathcal C^\infty([-1]\mathfrak g)\to\mathcal C^\infty([-1]\mathfrak h)$ is a quasi-isomorphism, i.e. it induces an isomorphism on cohomology. Does it follow that $f$ is an isomorphism?
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When I ask it this way, it sounds strongly like the answer should be "no": almost never is cohomology a complete invariant. For example, the cohomology in degree $1$ sees only the abelianizations of $\mathfrak g,\mathfrak h$. But on the other hand, research I'm doing on Lie algebroids suggests that the similar statement with "algebra" replaced by "algebroid" throughout should be true. I don't see a direct proof even in the "algebra" case, but I feel like there should be either a trivial counterexample or an easy argument in favor. In either case, though, and maybe because it's late at night, I'm stuck.


Which is all to say that secretly I care about algebroids, so if any of y'all know a good reference for the problem at that generality, please send it my way. But I will happily accept an answer just for algebras if one is provided.



Is a quasi-iso in Lie algebra cohomology necessarily an iso?
Here are some comments to your answer that I hope will be helpful (it's still sufficiently confusing that I might make some mistakes). Given an $L\_{\infty}$ algebra, we can define the "Koszul dual" Chevalley $\it{chains}$ $C\_\*(L)$(for several reasons it's more natural to think of the Chevalley chains rather than cochains). The ordinary notion of a quasi-isomorphism of $L\_{\infty}$ is a morphism of Chevalley complexes whose first Taylor coefficient induces a quasi-isomorphism of complexes. We could ask what happens if we declare two Lie algebras weakly equivalent if there is a quism of Chevalley complexes? We have seen that this notion can differ from the standard one. Another example is when $g\_1=sl\_2(\mathbb{C})$ and $g\_2=\mathbb{C}[2]$ an abelian Lie algebra concentrated in degree 2. From this point of view, your notion is a very natural relaxing of the usual notion of quasi-isomorphism --- it's the notion of weak equivalence transported from Koszul duality. It's worth noting that the two notions will coincide if your dg-Lie algebra is concentrated in positive degree (or more generally with some nilpotency hypotheses on the action of the degree zero piece $g^0$ on your dg lie algebra.) This explains Trial's answer above. 


It's useful to have the rational homotopy picture in mind. Given a 1-connected CW complex Xwhich is finite in each degree, you can construct a Lie model L for X whose $H\_{\*}(L)=\pi\_{\*}(\Omega(M)) $ with Whitehead bracket. One gets chains on your space by taking $C\_{\*}(L)$. To go back, you have something known as the cobar construction which spits out $C\_{\*}(\Omega(M))$, which is U(L), the universal enveloping algebra of L. Again, things start to break down in the non-simply connected case if you don't assume some hypothesis about the action of $\pi\_1$ on the higher homotopy groups. From this point of view, the comment at the end of the last paragraph is a reflection of the fact that in these instances, when you have an isomorphism on homology groups, you have an isomorphism on homotopy groups.


Finally when thinking about the idea that the more relaxed notion gives you a derived equivalence, you must define the derived category properly. Let's think about the abelian case. The most classical case of Koszul duality says that given a finite dimensional graded vector space, $D^{+}(SV) \cong D^{+}(\wedge(V^\*))$. Here we are considering the derived category of modules which are bounded below. It's pretty easy to see that this cannot be extended an equivalence of unbounded derived categories. 


The right notion which makes the whole thing work for unbounded derived categories can be found in Positselski's works and is called the coderived category. Let g be a Lie algebra over $\mathbb{C}$. Then the equivalence between the derived category of modules over U(g) and the coderived category of co-modules over it's Chevalley complex $C\_{∗}(g)$ in which $M \to C\_{∗}(g,M)$. Under the assumption g is finite, we can consider these co-modules to be modules over C\*(g)(now looking at cochains again) and look at the corresponding localization of the category of dg-modules over C\*(g). What we see is that we are localizing at a smaller set of weak equivalences than quasi-isomorphisms.


