Corollary 1

a) In the diagram (18), all the morphisms are homotopisms.

b) Let $\omega \in$ Alt($\mathfrak{g}$). Then $\omega$ belongs to Alt($\mathfrak{g}$)$^G$ if and only if $d\omega = 0$ and $d(i(\xi )\omega ) = 0$ for all $\xi \in \mathfrak{g}$. The differential of the complex Alt($\mathfrak{g}$)$^G$ is zero.

c) The graded vector space $H(\Omega (G))$ is isomorphic to Alt($\mathfrak{g}$)$^G$.

The theorem, applied to the action of G on G by left translations (resp. to the action $((g, h);x) \rightarrow gxh^{-1}$ of $G\times G$ on G) implies that the canonical injection $\Omega (G)^G\rightarrow \Omega (G)$ (resp. $^G\Omega (G)^G\rightarrow \Omega (G)$) is a homotopism; in view of Algèbre, Chap. X, p. 34, Cor., assertion a) follows.

We prove b). By Prop. 51 of Chap. III, §3, no. 14, we have $d\alpha =-d\alpha$, that is $d\alpha = 0$, for every differential form $\alpha$ on G that is left and right invariant. Thus, if $\omega \in$ Alt($\mathfrak{g}$)$^G$, then $d\omega = 0$, and consequently $d(i(\xi )\omega ) =$ $\theta (\xi )\omega -i(\xi )d\omega = 0$. Conversely, if $d\omega = 0$ and $d(i(\xi )d\omega ) = 0$, then $\theta (\xi )\omega = 0$.

Assertion c) follows from a) and b).