Lemma 76.44.1. Let $\mathcal{P}$ be a property of morphisms of schemes which is étale local on the target. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Consider commutative diagrams
\[ \xymatrix{ X \times _ Y V \ar[d] \ar[r] & V \ar[d] \\ X \ar[r]^ f & Y } \]
where $V$ is a scheme and $V \to Y$ is étale. The following are equivalent
for any diagram as above the projection $X \times _ Y V \to V$ has property $\mathcal{P}$, and
for some diagram as above with $V \to Y$ surjective the projection $X \times _ Y V \to V$ has property $\mathcal{P}$.
If $X$ and $Y$ are representable, then this is also equivalent to $f$ (as a morphism of schemes) having property $\mathcal{P}$.
Proof.
Let us prove the equivalence of (1) and (2). The implication (1) $\Rightarrow $ (2) is immediate. Assume
\[ \xymatrix{ X \times _ Y V \ar[d] \ar[r] & V \ar[d] \\ X \ar[r]^ f & Y } \quad \quad \xymatrix{ X \times _ Y V' \ar[d] \ar[r] & V' \ar[d] \\ X \ar[r]^ f & Y } \]
are two diagrams as in the lemma. Assume $V \to Y$ is surjective and $X \times _ Y V \to V$ has property $\mathcal{P}$. To show that (2) implies (1) we have to prove that $X \times _ Y V' \to V'$ has $\mathcal{P}$. To do this consider the diagram
\[ \xymatrix{ X \times _ Y V \ar[d] & (X \times _ Y V) \times _ X (X \times _ Y V') \ar[l] \ar[d] \ar[r] & X \times _ Y V' \ar[d] \\ V & V \times _ Y V' \ar[l] \ar[r] & V' } \]
By our assumption that $\mathcal{P}$ is étale local on the source, we see that $\mathcal{P}$ is preserved under étale base change, see Descent, Lemma 35.22.2. Hence if the left vertical arrow has $\mathcal{P}$ the so does the middle vertical arrow. Since $U \times _ X U' \to U'$ is surjective and étale (hence defines an étale covering of $U'$) this implies (as $\mathcal{P}$ is assumed local for the étale topology on the target) that the left vertical arrow has $\mathcal{P}$.
If $X$ and $Y$ are representable, then we can take $\text{id}_ Y : Y \to Y$ as our étale covering to see the final statement of the lemma is true.
$\square$
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