Lemma 73.9.2. Any fppf covering is an fpqc covering, and a fortiori, any syntomic, smooth, étale or Zariski covering is an fpqc covering.

**Proof.**
We will show that an fppf covering is an fpqc covering, and then the rest follows from Lemma 73.7.2. Let $\{ f_ i : U_ i \to U\} _{i \in I}$ be an fppf covering of algebraic spaces over $S$. By definition this means that the $f_ i$ are flat which checks the first condition of Definition 73.9.1. To check the second, let $V \to U$ be a morphism with $V$ affine. We may choose an étale covering $\{ V_{ij} \to V \times _ U U_ i\} $ with $V_{ij}$ affine. Then the compositions $f_{ij} : V_{ij} \to V \times _ U U_ i \to V$ are flat and locally of finite presentation as compositions of such (Morphisms of Spaces, Lemmas 67.28.2, 67.30.3, 67.39.7, and 67.39.8). Hence these morphisms are open (Morphisms of Spaces, Lemma 67.30.6) and we see that $|V| = \bigcup _{i \in I} \bigcup _{j \in J_ i} f_{ij}(|V_{ij}|)$ is an open covering of $|V|$. Since $|V|$ is quasi-compact, this covering has a finite refinement. Say $V_{i_1j_1}, \ldots , V_{i_ Nj_ N}$ do the job. Then $\{ V_{i_ kj_ k} \to V\} _{k = 1, \ldots , N}$ is a standard fpqc covering of $V$ refinining the family $\{ U_ i \times _ U V \to V\} $. This finishes the proof.
$\square$

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