Lemma 88.4.2. Let $f: R \to S$ be a ring map in $\widehat{\mathcal{C}}_\Lambda$. The following are equivalent

1. $f$ is surjective,

2. the map $\mathfrak m_ R/\mathfrak m_ R^2 \to \mathfrak m_ S/\mathfrak m_ S^2$ is surjective, and

3. the map $\mathfrak m_ R/(\mathfrak m_\Lambda R + \mathfrak m_ R^2) \to \mathfrak m_ S/(\mathfrak m_\Lambda S + \mathfrak m_ S^2)$ is surjective.

Proof. Note that for $n \geq 2$ we have the equality of relative cotangent spaces

$\mathfrak m_ R/(\mathfrak m_\Lambda R + \mathfrak m_ R^2) = \mathfrak m_{R_ n}/(\mathfrak m_\Lambda R_ n + \mathfrak m_{R_ n}^2)$

and similarly for $S$. Hence by Lemma 88.3.5 we see that $R_ n \to S_ n$ is surjective for all $n$. Now let $K_ n$ be the kernel of $R_ n \to S_ n$. Then the sequences

$0 \to K_ n \to R_ n \to S_ n \to 0$

form an exact sequence of directed inverse systems. The system $(K_ n)$ is Mittag-Leffler since each $K_ n$ is Artinian. Hence by Algebra, Lemma 10.85.4 taking limits preserves exactness. So $\mathop{\mathrm{lim}}\nolimits R_ n \to \mathop{\mathrm{lim}}\nolimits S_ n$ is surjective, i.e., $f$ is surjective. $\square$

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