Let $\mathfrak{g}$ be a Lie algebra. How can I show that $H^1(\mathfrak{g})\cong \mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$, the first real cohomology space of $\mathfrak{g}$?
My try: Let $c:\mathfrak{g}\to \mathbb{R}$ be a one-cocycle. Then $c$ is a linear map such that $c([X,Y])=0$ for all Lie algebra elements $X,Y$, which implies that $Ker(c)=[\mathfrak{g},\mathfrak{g}]$. So $c$ induces a linear map $\bar{c}:\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]\to \mathbb{R}$. Now how can I get the result?
I know that I need to define an epimorphism $\psi:\mathfrak{g}\to H^1 (\mathfrak{g})$ with $Ker(\psi)=[\mathfrak{g},\mathfrak{g}]$ but I don't know how to do it.
By definition of Lie algebra cohomology, with the trivial $1$-dimensional module $K$, we have $$ H^1(\mathfrak{g},K)=\{\omega\in \mathfrak{g}^*\mid \omega([\mathfrak{g},\mathfrak{g}])=0\}\cong (\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^*, $$ see here, page $47$. This is already your result. For a finite-dimensional vector space $V$ we have $V^*\cong V$ then. So you need to assume that $\mathfrak{g}$ is finite-dimensional.