**Proof.**
See More on Morphisms, Lemma 37.6.8 for a more general statement and proof. What follows is a sketch of the proof in the current case.

Firstly, one checks both properties are local on the source and the target. This we may assume that $S$ and $X$ are affine. Say $X = \mathop{\mathrm{Spec}}(B)$ and $S = \mathop{\mathrm{Spec}}(R)$. Say $T = \mathop{\mathrm{Spec}}(C)$. Let $J$ be the square-zero ideal of $C$ with $T_0 = \mathop{\mathrm{Spec}}(C/J)$. Assume that we are given the diagram

\[ \xymatrix{ & B \ar[d]^\phi \ar[rd]^{\bar{\phi }} & \\ R \ar[r] \ar[ur] & C \ar[r] & C/J } \]

Secondly, one checks that the association $\phi ' \mapsto \phi ' - \phi $ gives a bijection between the set of liftings of $\bar{\phi }$ and the module $\text{Der}_ R(B, J)$. Thus, we obtain the implication (1) $\Rightarrow $ (2) via the description of unramified morphisms having trivial module of differentials, see Theorem 41.4.1.

To obtain the reverse implication, consider the surjection $q : C = (B \otimes _ R B)/I^2 \to B = C/J$ defined by the square zero ideal $J = I/I^2$ where $I$ is the kernel of the multiplication map $B \otimes _ R B \to B$. We already have a lifting $B \to C$ defined by, say, $b \mapsto b \otimes 1$. Thus, by the same reasoning as above, we obtain a bijective correspondence between liftings of $\text{id} : B \to C/J$ and $\text{Der}_ R(B, J)$. The hypothesis therefore implies that the latter module is trivial. But we know that $J \cong \Omega _{B/R}$. Thus, $B/R$ is unramified.
$\square$

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