Let $\varphi:\mathfrak{g}\to\mathfrak{q}$ be a morphism of finite-dimensional Lie algebras over a field $K$. Define $\varphi^\vee:\mathfrak{q}^\vee\to\mathfrak{g}^\vee$, $l\mapsto l\circ\varphi$, where $\mathfrak{g}^\vee$ is the $K$-vector space of all Lie algebra morphisms from $\mathfrak{g}$ to $K$. Let $0\to\mathfrak{h}\xrightarrow{\iota}\mathfrak{g}\xrightarrow{\varphi}\mathfrak{q}\to 0$ be a short exact sequence of Lie algebras.
Question: Is the "dual" sequence of vector spaces $0\to\mathfrak{q}^\vee\xrightarrow{\varphi^\vee}\mathfrak{g}^\vee\xrightarrow{\iota^\vee}\mathfrak{h}^\vee\to 0$ also exact?
My thoughts: I was able to prove that $\varphi^\vee$ is injective and that $\text{im}(\varphi^\vee)\subset\text{ker}(\iota^\vee)$. To prove the other inclusion, I tried proving that one can identify $\text{ker}(\iota^\vee)$ with a subspace of $\text{im}(\iota)^\vee$ and that one can identify $\text{ker}(\varphi)^\vee$ with a subspace of $\text{im}(\varphi^\vee)$. I was able to prove the former of these identifications.
I have trouble showing the rest, i.e. showing that $\iota^\vee$ is surjective and that $\text{ker}(\iota^\vee)\subset\text{im}(\varphi^\vee)$ (or that there exists an injective linear map $\text{ker}(\varphi)^\vee\hookrightarrow\text{im}(\varphi^\vee)$). For surjectivity, I know that $\iota$ has a linear retraction but of course it need not respect the brackets. The vector space dual version of this statement uses the fact that every vector subspace has a complement but of course we cannot assume that every Lie subalgebra has a complementary Lie algebra. Any ideas?
Edit: I have been able to prove everything except the surjectivity of $\iota^\vee$.