**Proof.**
Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective étale morphism $U' \to X' \times _{Y'} V'$. Set $V = Y \times _{Y'} V'$ and $U = X \times _{X'} U'$. Then for étale local properties of morphisms we can reduce to the morphism of thickenings of schemes $(U \subset U') \to (V \subset V')$ and apply More on Morphisms, Lemma 37.3.3. This proves (2), (3), (4), (5), and (6).

The properties of morphisms in (1), (7), (8), (9), (10) are stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Spaces, Lemma 65.12.3, and Morphisms of Spaces, Lemmas 67.40.3, 67.45.5, and 67.10.5.

The interesting direction in (1), (7), (8), (9), (10) is 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 $Y \subset Y'$ is a first order thickening, see discussion on finite order thickenings above.

Proof of (1). Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Set $V = Y \times _{Y'} V'$, $U' = X' \times _{Y'} V'$ and $U = X \times _ Y V$. Then $U \to V$ is a closed immersion, which implies that $U$ is a scheme, which in turn implies that $U'$ is a scheme (Lemma 76.9.5). Thus we can apply the lemma in the case of schemes (More on Morphisms, Lemma 37.3.3) to $(U \subset U') \to (V \subset V')$ to conclude.

Proof of (7). Follows by combining (2) with results of Lemma 76.10.1 and the fact that proper equals quasi-compact $+$ separated $+$ locally of finite type $+$ universally closed.

Proof of (8). Follows by combining (2) with results of Lemma 76.10.1 and using the fact that finite equals integral $+$ locally of finite type (Morphisms, Lemma 29.44.4).

Proof of (9). As $f$ is a monomorphism we have $X = X \times _ Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ Y X \subset X' \times _{Y'} X')$. We conclude $X' \to X' \times _{Y'} X'$ is a closed immersion by (1). In fact, it is a first order thickening as the ideal defining the closed immersion $X' \to X' \times _{Y'} X'$ is contained in the pullback of the ideal $\mathcal{I} \subset \mathcal{O}_{Y'}$ cutting out $Y$ in $Y'$. Indeed, $X = X \times _ Y X = (X' \times _{Y'} X') \times _{Y'} Y$ is contained in $X'$. The conormal sheaf of the closed immersion $\Delta : X' \to X' \times _{Y'} X'$ is equal to $\Omega _{X'/Y'}$ (this is the analogue of Morphisms, Lemma 29.32.7 for algebraic spaces and follows either by étale localization or by combining Lemmas 76.7.11 and 76.7.13; some details omitted). Thus it suffices to show that $\Omega _{X'/Y'} = 0$ which follows from (5) and the corresponding statement for $X/Y$.

Proof of (10). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a closed immersion, see Morphisms of Spaces, Remark 67.12.4. Let $V' \subset Y'$ be the open subspace whose underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$. Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (1). This finishes the proof.
$\square$

## Comments (2)

Comment #1564 by Matthew Emerton on

Comment #1582 by Johan on