Proposition 12

Let B be a K-algebra, $u': L\rightarrow B$ a K-algebra homomorphism, and $v': \Gamma \rightarrow B^*$ a group homomorphism. We assume satisfied the relations

for $\alpha \in L^*,\beta \in L$, and $\gamma \in \Gamma$. Then there exists a unique algebra homomorphism $f$ from $\mathbf{A}[\mathscr{E}; L]$ to B such that we have $u'=f\circ u$ and $v'=f\circ v$.

To prove the uniqueness of $f$, observe that the vector space $\mathbf{A}[\mathscr{E}; L]$ over the field K is generated by the set of elements of the form $[\beta ;\gamma ] =u(\beta )v(\gamma )$. If the homomorphism $f':\mathbf{A}[\mathscr{E}; L]\rightarrow B$ satisfies $f'\circ u=u'$ and $f'\circ v=v'$, then it sends $[\beta ;\gamma ]$ to $u'(\beta )v'(\gamma )$ and therefore coincides with $f$.

By formula (29), we have

for $\alpha$ in $L^*,\beta$ in L, and $\gamma$ in Γ. By the definition of E, there consequently exists a mapping $f_0: E\rightarrow B$ such that $f_0([\beta ;\gamma ]) =u'(\beta )v'(\gamma )$. By formulas (23) and (30), we have $f_0(xx') =f_0(x)f_0(x')$ for $x$ and $x'$ in E. The restriction of $f_0$ to $E_g$ is K-linear for every element $g$ of G; consequently, there exists a unique K-linear mapping $f$ from $\mathbf{A}[\mathscr{E}; L]$ to B that coincides with $f_0$ on E. The mapping $f$ is an algebra morphism that satisfies $u'=f\circ u$ and $v'=f\circ v$.