**Proof.**
As the problem is local on $X'$ we may assume that $X, X', S, S'$ are affine schemes. Say $S' = \mathop{\mathrm{Spec}}(A')$, $X' = \mathop{\mathrm{Spec}}(B')$, $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ with $A = A'/I$ and $B = B'/J$ for some square zero ideals. Then we obtain the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } \]

with exact rows. The canonical map of the lemma is the map

\[ I \otimes _ A B = I \otimes _{A'} B' \longrightarrow J. \]

The assumption that $f$ is flat signifies that $A \to B$ is flat.

Assume (1). Then $A' \to B'$ is flat and $J = IB'$. Flatness implies $\text{Tor}_1^{A'}(B', A) = 0$ (see Algebra, Lemma 10.75.8). This means $I \otimes _{A'} B' \to B'$ is injective (see Algebra, Remark 10.75.9). Hence we see that $I \otimes _ A B \to J$ is an isomorphism.

Assume (2). Then it follows that $J = IB'$, so that $X = S \times _{S'} X'$. Moreover, we get $\text{Tor}_1^{A'}(B', A'/I) = 0$ by reversing the implications in the previous paragraph. Hence $B'$ is flat over $A'$ by Algebra, Lemma 10.99.8.
$\square$

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