Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Let $K$ be a closed subgroup of $G$ with corresponding Lie subalgebra $\mathfrak{k}$. Let $V$ be a $\left(\mathfrak{g},K\right)$-module. Then, I want to show that the relative Lie algebra cohomology $H^q(\mathfrak{g},K;V)$ is equal to $Ext_{\left(\mathfrak{g},K\right)}^q\left(\mathbb{C},V\right)$, where $\mathbb{C}$ is the trivial $(\mathfrak{g},K)$-module.
A hint that I got from my advisor (I am not sure if I remember it accurately): Take a special projective resolution of $\mathbb{C}$ of the form
$\cdots \rightarrow \bigwedge^k(\mathfrak{g}/\mathfrak{k}) \rightarrow \bigwedge^{k-1}(\mathfrak{g}/\mathfrak{k}) \rightarrow \cdots \rightarrow \bigwedge^1(\mathfrak{g}/\mathfrak{k}) \rightarrow \mathbb{C}$
Now taking homomorphisms into $V$ gives the same complex that comes in the definition of Relative Lie algebra cohomology, and hence these groups are the same.
Can somebody make it more precise? What are the maps in the above complex?