Let $R$ be a commutative ring, $G$ a finite group, $P$ be a projective, finitely generated, $R[G]$-module, let $P_0$ be $P$ considered as a $R$-module. Serre states in Linear Representation of finite groups (14.4(Lemma 20)) that it can be shown that for any $R[G]$ morphism $v : P \rightarrow R[G] \otimes_RP_0 $ there exists $u\in \operatorname{End}_R(P_0)$ such that :
$$ v(x) = \sum_{s\in G} s \otimes u(s^{-1}x) $$
Now obviously $R[G] \otimes_RP_0$ is projective and finitely generated but I do not see how to proceed. Any help or pointer toward a proof of the precedent statement would be appreciated.
I think this is a valid proof. First $G$ is a generating set of $R[G]$, so indeed any $v\in \operatorname{Mor}_{R[G]}(P, R[G] \otimes_RP_0)$ can be decomposed through :
$$\forall x \in P, v(x) = \sum_{s\in G} s\otimes u_s(x)$$ with $u_s \in \operatorname{End}_{R}(P_0)$. Then since $v$ is a $R[G]$ morphism we have $\forall t\in G, \forall x \in P, v(tx) = tv(x)$, thus :
$$ 1_G \otimes u_1(tx) + t\otimes u_t(tx) + \sum_{s\in G-\{1_G, t\}} s\otimes u_s(tx) $$ $$= t \otimes u_1(x) + t^2\otimes u_t(x) + \sum_{s\in G-\{1_G, t\}} ts\otimes u_s(x) $$
By identification of components along the generating set $G$ we get that : $$t \otimes u_1(x) = t\otimes u_t(tx)$$ $$\implies u_1(x) = u_t(tx)$$ $$\implies u_1(t^{-1}x) = u_t(x)$$ $$\implies v(x) = \sum_{s\in G} s\otimes u_1(s^{-1}x)$$