**Proof.**
Assume (3). Let $C \to B$ be a ring map in $\mathcal{C}_\Lambda $ such that $C \to A$ is surjective. Then $C \to A/(\mathfrak m_\Lambda A + \mathfrak m_ A^2)$ is surjective too. We conclude that $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2)$ is surjective by our assumption. Hence $C \to B$ is surjective by applying Lemma 89.3.5 (2 times).

Assume (1). Let $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2)$ be a morphism of $\mathcal{C}_\Lambda $ such that $C \to A/(\mathfrak m_\Lambda A + \mathfrak m_ A^2)$ is surjective. Set $C' = C \times _{B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2)} B$ which is an object of $\mathcal{C}_\Lambda $ by Lemma 89.3.8. Note that $C' \to A/(\mathfrak m_\Lambda A + \mathfrak m_ A^2)$ is still surjective, hence $C' \to A$ is surjective by Lemma 89.3.5. Thus $C' \to B$ is surjective by our assumption. This implies that $C' \to B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2)$ is surjective, which implies by the construction of $C'$ that $C \to B/(\mathfrak m_\Lambda B + \mathfrak m_ B^2)$ is surjective.

In the first paragraph we proved (3) $\Rightarrow $ (1) and in the second paragraph we proved (1) $\Rightarrow $ (3). The equivalence of (2) and (3) is a special case of the equivalence of (1) and (3), hence we are done.
$\square$

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