Lemma 64.15.6. With assumptions as in Lemma 64.15.5, let $u\in \text{End}_{A[G]}(P)$ and $v\in \text{End}_{\Lambda [G]}(M)$. Then
Sketch of proof. Reduce to the case $P=A[G]$. In that case, $u$ is right multiplication by some element $a = \sum a_ gg$ of $A[G]$, which we write $u = R_ a$. There is an isomorphism of $\Lambda [G]$-modules
where $\left(A[G]\otimes _ A M\right)'$ has the module structure given by the left $G$-action, together with the $\Lambda $-linearity on $M$. This transport of structure changes $u \otimes v$ into $\sum _ ga_ gR_ g \otimes g^{-1}v$. In other words,
Working out explicitly both sides of the equation, we have to show
This is done by showing that
by reducing to $M=\Lambda $. $\square$
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