**Proof.**
We omit the proof that (1) implies (2), and that (2) implies (3).

Assume (3). Choose a surjective étale morphism $V \to Y$. We are going to show that $V \times _ Y X \to V$ is a universally open morphism of algebraic spaces. Let $Z \to V$ be a morphism from an algebraic space to $V$. Let $W \to Z$ be a surjective étale morphism where $W = \coprod W_ i$ is a disjoint union of affine schemes, see Properties of Spaces, Lemma 66.6.1. Then we have the following commutative diagram

\[ \xymatrix{ \coprod _ i |W_ i \times _ Y X| \ar@{=}[r] \ar[d] & |W \times _ Y X| \ar[r] \ar[d] & |Z \times _ Y X| \ar[d] \ar@{=}[r] & |Z \times _ V (V \times _ Y X)| \ar[ld] \\ \coprod |W_ i| \ar@{=}[r] & |W| \ar[r] & |Z| } \]

We have to show the south-east arrow is open. The middle horizontal arrows are surjective and open (Properties of Spaces, Lemma 66.16.7). By assumption (3), and the fact that $W_ i$ is affine we see that the left vertical arrows are open. Hence it follows that the right vertical arrow is open.

Assume $V \to Y$ is as in (4). We will show that $f$ is universally open. Let $Z \to Y$ be a morphism of algebraic spaces. Consider the diagram

\[ \xymatrix{ |(V \times _ Y Z) \times _ V (V \times _ Y X)| \ar@{=}[r] \ar[rd] & |V \times _ Y X| \ar[r] \ar[d] & |Z \times _ Y X| \ar[d] \\ & |V \times _ Y Z| \ar[r] & |Z| } \]

The south-west arrow is open by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic spaces are étale (see Properties of Spaces, Lemma 66.16.7). It follows that the right vertical arrow is open.

Of course (1) implies (5) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (5). Then for any $Z \to Y$ we get a corresponding Zariski covering $Z = \bigcup Z_ i$ such that the base change of $f$ to $Z_ i$ is open. By a simple topological argument this implies that $Z \times _ Y X \to Z$ is open. Hence (1) holds.
$\square$

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