First cohomology with trivial coefficients

A linear map $\lambda :\mathfrak{g}\to k$ is a derivation precisely when $\lambda ([x,y])=x\cdot \lambda (y)-y\cdot \lambda (x)=0$. Thus $Z^1(\mathfrak{g},k)$ is exactly the space of linear forms vanishing on $[\mathfrak{g},\mathfrak{g}]$.

Every inner derivation into the trivial module has the form $x\mapsto x\cdot a=0$, so $B^1(\mathfrak{g},k)=0$ and $H^1=Z^1$.

Put $I=[\mathfrak{g},\mathfrak{g}]$. If $\lambda$ is in the space from step 1.1, define $\bar{\lambda }(x+I)=\lambda (x)$. This is well-defined because $x+I=y+I$ implies $x-y\in I$ and hence $\lambda (x-y)=0$; it is plainly linear and satisfies $\lambda =\bar{\lambda }\circ q$. Conversely every linear form on $\mathfrak{g}\mathrm{/}I$ pulls back along the linear map $q$ from [L2] to a form vanishing on $I$. These constructions are linear and inverse, proving the displayed natural isomorphism together with steps 1.1–1.2. If the abelianization is zero, both sides are zero; no basis or choice is used.