Lemma 10.118.4. Let $R \to S$ be a ring map. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $S$-modules. Then

$U(R \to S, M_1) \cap U(R \to S, M_3) \subset U(R \to S, M_2).$

Proof. Let $u \in U(R \to S, M_1) \cap U(R \to S, M_3)$. Choose $f_1, f_3 \in R$ such that $u \in D(f_1)$, $u \in D(f_3)$ and such that (10.118.3.1) holds for $f_1$ and $M_1$ and for $f_3$ and $M_3$. Then set $f = f_1f_3$. Then $u \in D(f)$ and (10.118.3.1) holds for $f$ and both $M_1$ and $M_3$. An extension of free modules is free, and an extension of finitely presented modules is finitely presented (Lemma 10.5.3). Hence we see that (10.118.3.1) holds for $f$ and $M_2$. Thus $u \in U(R \to S, M_2)$ and we win. $\square$

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