**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 37.3.4. This proves (1), (4), and (5).

The properties in (2), (3), (6), (7), and (8) 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. Hence in each case we need only to prove that if $f$ has the desired property, so does $f'$.

Case (2) follows from case (5) of Lemma 76.10.1 and the fact that the finite morphisms are precisely the integral morphisms that are locally of finite type (Morphisms of Spaces, Lemma 67.45.6).

Case (3). This follows immediately from Limits of Spaces, Lemma 70.15.5.

Proof of (6). 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 $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$ is a closed immersion by (3). In fact $\Delta _{X'/Y'}$ induces a bijection $|X'| \to |X' \times _{Y'} X'|$, hence $\Delta _{X'/Y'}$ is a thickening. On the other hand $\Delta _{X'/Y'}$ is locally of finite presentation by Morphisms of Spaces, Lemma 67.28.10. In other words, $\Delta _{X'/Y'}(X')$ is cut out by a quasi-coherent sheaf of ideals $\mathcal{J} \subset \mathcal{O}_{X' \times _{Y'} X'}$ of finite type. Since $\Omega _{X'/Y'} = 0$ by (5) we see that the conormal sheaf of $X' \to X' \times _{Y'} X'$ is zero. (The conormal sheaf of the closed immersion $\Delta _{X'/Y'}$ 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.) In other words, $\mathcal{J}/\mathcal{J}^2 = 0$. This implies $\Delta _{X'/Y'}$ is an isomorphism, for example by Algebra, Lemma 10.21.5.

Proof of (7). 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 (3).

Case (8) follows from Lemma 76.10.1 and the definition of proper morphisms as being the quasi-compact, universally closed, and separated morphisms that are locally of finite type.
$\square$

## Comments (1)

Comment #1600 by Matthew Emerton on