Example 2

It follows from the preceding that $H^1(G, M)$ is isomorphic to the $\mathbf{Z}$-module $Z^1(G, M)/B^1(G, M)$, where $Z^1(G, M)$ is the $\mathbf{Z}$-module of mappings $f$ of $G$ into $M$ satisfying:

and $B^1(G, M)$ is the sub-$\mathbf{Z}$-module of $Z^1(G, M)$ consisting of the $f$ for which there exists an element $m$ of $M$ such that:

One sometimes says that $Z^1(G, M)$ is the $\mathbf{Z}$-module of crossed homomorphisms of $G$ into $M$, and $B^1(G, M)$ the submodule of principal crossed homomorphisms.

Let us denote by $\iota:G\to \operatorname{Aut}(M)$ the homomorphism deduced from the action of $G$; consider the external semidirect product $M\times_{\iota}G$ and the extension $\xi_{\iota}:M\times_{\iota}G\to G$ (I, p. 64). Let $e:G\to M\times_{\iota}G$ be a mapping such that $\rho\circ e=1_G$; one has $e=(f,1_G)$, where $f\in C^1(G,M)$. In order that $e$ be a homomorphism (that is to say, a section of the extension $\xi_{\iota}$), it is necessary and sufficient that $f\in Z^1(G,M)$. In order that two sections of $\xi_{\iota}$ be conjugate by an element of $i(M)$, it is necessary and sufficient that the corresponding crossed homomorphisms have the same class in $H^1(G,M)$.

When $G$ operates trivially on $M$, one has $B^1(G,M)=0$ and $H^1(G,M)$ is isomorphic to the $\mathbf{Z}$-module of group homomorphisms of $G$ into $M$.