**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) – (20) 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, 29.20.13, 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. 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. (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. 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). By Algebra, Lemma 10.126.11 we see that $A' \to B'$ is an isomorphism, 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 $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$. Observe that $X' \times _{Y'} X' \to S'$ is locally of finite type. 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). Follows from the fact that $f'$ is locally of finite type (by Morphisms, Lemma 29.15.8) and that quasi-finiteness of a morphism of finite type can be checked on fibres, see Morphisms, Lemma 29.20.6.

Ad (12). Follows from general remark (II) and the fact that $f'$ is locally of finite type (Morphisms, Lemma 29.15.8).

Ad (13). Immediate from general remark (II). See also Lemma 37.3.1 for a more general statement.

Ad (14). Assume $f$ is syntomic. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is syntomic by Morphisms, Lemma 29.30.11.

Ad (15). Assume $f$ is smooth. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is smooth by Morphisms, Lemma 29.34.3.

Ad (16). Assume $f$ unramified. By Morphisms, Lemma 29.15.8 $f'$ is locally of finite type. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is unramified by Morphisms, Lemma 29.35.12.

Ad (17). Assume $f$ étale. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. By general remark (III) $f'$ is flat. The fibres of $f'$ are the fibres of $f$. Hence $f'$ is étale by Morphisms, Lemma 29.36.8.

Ad (18). This follows from a combination of (6), the fact that $f$ is locally of finite type (Morphisms, Lemma 29.15.8), (4), and (5).

Ad (19). Combine (5), (10), Morphisms, Lemma 29.44.7, the fact that $f$ is locally of finite type (Morphisms, Lemma 29.15.8), and Morphisms, Lemma 29.44.4.

Ad (20). Assume $f$ finite locally free. By (19) we see that $f'$ is finite. By general remark (III) $f'$ is flat. By Morphisms, Lemma 29.21.11 $f'$ is locally of finite presentation. Hence $f'$ is finite locally free by Morphisms, Lemma 29.48.2.
$\square$

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