Lemma 10.18.4. Let $\varphi : R \to S$ be a ring map. Assume $R$ and $S$ are local rings. The following are equivalent:
$\varphi $ is a local ring map,
$\varphi (\mathfrak m_ R) \subset \mathfrak m_ S$, and
$\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$.
For any $x \in R$, if $\varphi (x)$ is invertible in $S$, then $x$ is invertible in $R$.
Proof. Conditions (1) and (2) are equivalent by definition. If (3) holds then (2) holds. Conversely, if (2) holds, then $\varphi ^{-1}(\mathfrak m_ S)$ is a prime ideal containing the maximal ideal $\mathfrak m_ R$, hence $\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$. Finally, (4) is the contrapositive of (2) by Lemma 10.18.3. $\square$
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