Lemma 15.90.4. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces an isomorphism $R/fR \to R'/fR'$. The map $\mathop{\mathrm{Spec}}(R') \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective. For example, the map $\mathop{\mathrm{Spec}}(R^\wedge ) \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

**Proof.**
Recall that $\mathop{\mathrm{Spec}}(R) = V(f) \amalg D(f)$ where $V(f) = \mathop{\mathrm{Spec}}(R/fR)$ and $D(f) = \mathop{\mathrm{Spec}}(R_ f)$, see Algebra, Section 10.17 and especially Lemmas 10.17.7 and 10.17.6. Thus the lemma follows as the map $R \to R/fR$ factors through $R'$. The last statement is a special case of the first statement by Lemma 15.90.1.
$\square$

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