Lemma 46.3.13. Let $A$ be a ring. Let $F, G$ be module-valued functors on $\textit{Alg}_ A$. Let $\varphi : F \to G$ be a transformation of functors. Assume

$\varphi $ is additive,

for every $A$-algebra $B$ and $\xi \in F(B)$ and unit $u \in B^*$ we have $\varphi (u\xi ) = u\varphi (\xi )$ in $G(B)$, and

for any flat ring map $B \to B'$ we have $G(B) \otimes _ B B' = G(B')$.

Then $\varphi $ is a morphism of module-valued functors.

**Proof.**
Let $B$ be an $A$-algebra, $\xi \in F(B)$, and $b \in B$. We have to show that $\varphi (b \xi ) = b \varphi (\xi )$. Consider the ring map

\[ B \to B' = B[x, y, x^{-1}, y^{-1}]/(x + y - b). \]

This ring map is faithfully flat, hence $G(B) \subset G(B')$. On the other hand

\[ \varphi (b\xi ) = \varphi ((x + y)\xi ) = \varphi (x\xi ) + \varphi (y\xi ) = x\varphi (\xi ) + y\varphi (\xi ) = (x + y)\varphi (\xi ) = b\varphi (\xi ) \]

because $x, y$ are units in $B'$. Hence we win.
$\square$

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