**Proof.**
Proof of (1). Suppose that $y$ lies over the scheme $V$. We may think of $y$ as a morphism $(\mathit{Sch}/V)_{fppf} \to \mathcal{Y}$. By definition the $2$-fibre product $\mathcal{X} \times _\mathcal {Y} (\mathit{Sch}/V)_{fppf}$ is representable by an algebraic space $W$ and the morphism $W \to V$ is surjective, flat, and locally of finite presentation. Choose a scheme $U$ and a surjective étale morphism $U \to W$. Then $U \to V$ is also surjective, flat, and locally of finite presentation (see Morphisms of Spaces, Lemmas 67.39.7, 67.39.8, 67.5.4, 67.28.2, and 67.30.3). Hence $\{ U \to V\} $ is an fppf covering. Denote $x$ the object of $\mathcal{X}$ over $U$ corresponding to the $1$-morphism $(\mathit{Sch}/U)_{fppf} \to \mathcal{X}$. Then $\{ f(x) \to y\} $ is the desired fppf covering of $\mathcal{Y}$.

Proof of (2). Suppose that $y$ lies over the scheme $V$. We may think of $y$ as a morphism $(\mathit{Sch}/V)_{fppf} \to \mathcal{Y}$. By definition the $2$-fibre product $\mathcal{X} \times _\mathcal {Y} (\mathit{Sch}/V)_{fppf}$ is representable by an algebraic space $W$ and the morphism $W \to V$ is surjective and smooth. Choose a scheme $U$ and a surjective étale morphism $U \to W$. Then $U \to V$ is also surjective and smooth (see Morphisms of Spaces, Lemmas 67.39.6, 67.5.4, and 67.37.2). Hence $\{ U \to V\} $ is a smooth covering. By More on Morphisms, Lemma 37.38.7 there exists an étale covering $\{ V_ i \to V\} $ such that each $V_ i \to V$ factors through $U$. Denote $x_ i$ the object of $\mathcal{X}$ over $V_ i$ corresponding to the $1$-morphism

\[ (\mathit{Sch}/V_ i)_{fppf} \to (\mathit{Sch}/U)_{fppf} \to \mathcal{X}. \]

Then $\{ f(x_ i) \to y\} $ is the desired étale covering of $\mathcal{Y}$.
$\square$

## Comments (2)

Comment #5139 by Dario Weißmann on

Comment #5332 by Johan on

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