**Proof.**
The assumptions on $X$ and $Y$ mean that $f$ is the base change of $f'$ by $X \to X'$. The properties $\mathcal{P}$ listed in (1) – (23) above are all stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Schemes, Lemmas 26.18.2, 26.19.3, 26.21.12, and 26.23.5 and Morphisms, Lemmas 29.9.4, 29.10.4, 29.11.8, 29.15.4, 29.20.13, 29.21.4, 29.29.2, 29.30.4, 29.34.5, 29.35.5, 29.36.4, 29.41.5, 29.44.6, and 29.48.4.

The interesting direction in each case is therefore to assume that $f$ has the property and deduce that $f'$ has it too. By induction on the order of the thickening we may assume that $S \subset S'$ is a first order thickening, see discussion immediately following Definition 37.2.1. We make a couple of general remarks which we will use without further mention in the arguments below. (I) Let $W' \subset S'$ be an affine open and let $U' \subset X'$ and $V' \subset Y'$ be affine opens lying over $W'$ with $f'(U') \subset V'$. Let $W' = \mathop{\mathrm{Spec}}(R')$ and denote $I \subset R'$ be the ideal defining the closed subscheme $W' \cap S$. Say $U' = \mathop{\mathrm{Spec}}(B')$ and $V' = \mathop{\mathrm{Spec}}(A')$. Then we get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & IB' \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & IA' \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } \]

with exact rows. Moreover $IB' \cong I \otimes _ R B$, see proof of Lemma 37.10.1. (II) The morphisms $X \to X'$ and $Y \to Y'$ are universal homeomorphisms. Hence the topology of the maps $f$ and $f'$ (after any base change) is identical. (III) If $f$ is flat, then $f'$ is flat and $Y' \to S'$ is flat at every point in the image of $f'$, see Lemma 37.10.2.

Ad (1). This is general remark (III).

Ad (2). Assume $f$ is an isomorphism. By (III) we see that $Y' \to S'$ is flat. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $V \cong f^{-1}(V) = U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Thus we have a diagram as in the general remark (I) and moreover $IA \cong I \otimes _ R A$ because $R' \to A'$ is flat. Then $IB' \cong I \otimes _ R B \cong I \otimes _ R A \cong IA'$ and $A \cong B$. By the exactness of the rows in the diagram above we see that $A' \cong B'$, i.e., $U' \cong V'$. Thus $f'$ is an isomorphism.

Ad (3). Assume $f$ is an open immersion. Then $f$ is an isomorphism of $X$ with an open subscheme $V \subset Y$. Let $V' \subset Y'$ be the open subscheme whose underlying topological space is $V$. Then $f'$ is a map from $X'$ to $V'$ which is an isomorphism by (2). Hence $f'$ is an open immersion.

Ad (4). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (5). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (6). Note that $X \times _ Y X = Y \times _{Y'} (X' \times _{Y'} X')$ so that $X' \times _{Y'} X'$ is a thickening of $X \times _ Y X$. Hence the topology of the maps $\Delta _{X/Y}$ and $\Delta _{X'/Y'}$ matches and we win. See also Lemma 37.3.1 for a more general statement.

Ad (7). Assume $f$ is a monomorphism. Consider the diagonal morphism $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$. The base change of $\Delta _{X'/Y'}$ by $S \to S'$ is $\Delta _{X/Y}$ which is an isomorphism by assumption. By (2) we conclude that $\Delta _{X'/Y'}$ is an isomorphism.

Ad (8). This is clear. See also Lemma 37.3.1 for a more general statement.

Ad (9). Immediate from remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (10). Assume $f$ is affine. Choose an affine open $V' \subset Y'$ and set $U' = (f')^{-1}(V')$. Then $V = Y \cap V'$ is affine which implies that $U = Y \times _{Y'} U'$ is affine. By Lemma 37.2.3 we see that $U'$ is affine. Hence $f'$ is affine. See also Lemma 37.3.1 for a more general statement.

Ad (11). Via remark (I) comes down to proving $A' \to B'$ is of finite type if $A \to B$ is of finite type. Suppose that $x_1, \ldots , x_ n \in B'$ are elements whose images in $B$ generate $B$ as an $A$-algebra. Then $A'[x_1, \ldots , x_ n] \to B$ is surjective as both $A'[x_1, \ldots , x_ n] \to B$ is surjective and $I \otimes _ R A[x_1, \ldots , x_ n] \to I \otimes _ R B$ is surjective. See also Lemma 37.3.3 for a more general statement.

Ad (12). Follows from (11) and that quasi-finiteness of a morphism of finite type can be checked on fibres, see Morphisms, Lemma 29.20.6. See also Lemma 37.3.3 for a more general statement.

Ad (13). Via remark (I) comes down to proving $A' \to B'$ is of finite presentation if $A \to B$ is of finite presentation. We may assume that $B' = A'[x_1, \ldots , x_ n]/K'$ for some ideal $K'$ by (11). We get a short exact sequence

\[ 0 \to K' \to A'[x_1, \ldots , x_ n] \to B' \to 0 \]

As $B'$ is flat over $R'$ we see that $K' \otimes _{R'} R$ is the kernel of the surjection $A[x_1, \ldots , x_ n] \to B$. By assumption on $A \to B$ there exist finitely many $f'_1, \ldots , f'_ m \in K'$ whose images in $A[x_1, \ldots , x_ n]$ generate this kernel. Since $I$ is nilpotent we see that $f'_1, \ldots , f'_ m$ generate $K'$ by Nakayama's lemma, see Algebra, Lemma 10.20.1.

Ad (14). Follows from (11) and general remark (II). See also Lemma 37.3.3 for a more general statement.

Ad (15). Immediate from general remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (16). Assume $f$ is syntomic. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is syntomic by Morphisms, Lemma 29.30.11.

Ad (17). Assume $f$ is smooth. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is smooth by Morphisms, Lemma 29.34.3.

Ad (18). Assume $f$ unramified. By (11) $f'$ is locally of finite type and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is unramified by Morphisms, Lemma 29.35.12. See also Lemma 37.3.3 for a more general statement.

Ad (19). Assume $f$ étale. By (13) $f'$ is locally of finite presentation, by general remark (III) $f'$ is flat, and the fibres of $f'$ are the fibres of $f$. Hence $f'$ is étale by Morphisms, Lemma 29.36.8.

Ad (20). This follows from a combination of (6), (11), (4), and (5). See also Lemma 37.3.3 for a more general statement.

Ad (21). Combine (5) and (10) with Morphisms, Lemma 29.44.7. See also Lemma 37.3.1 for a more general statement.

Ad (22). Combine (21), and (11) with Morphisms, Lemma 29.44.4. See also Lemma 37.3.3 for a more general statement.

Ad (23). Assume $f$ finite locally free. By (22) we see that $f'$ is finite, by general remark (III) $f'$ is flat, and by (13) $f'$ is locally of finite presentation. Hence $f'$ is finite locally free by Morphisms, Lemma 29.48.2.
$\square$

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