**Proof.**
Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks with $\mathcal{Z}$ quasi-compact. By Properties of Stacks, Lemma 100.6.2 there exists a quasi-compact scheme $U$ and a surjective smooth morphism $U \to \mathcal{Z}$. Since $f$ is representable by algebraic spaces and quasi-compact we see by definition that $U \times _\mathcal {Y} \mathcal{X}$ is an algebraic space, and that $U \times _\mathcal {Y} \mathcal{X} \to U$ is quasi-compact. Hence $U \times _ Y X$ is a quasi-compact algebraic space. The morphism $U \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is smooth and surjective (as the base change of the smooth and surjective morphism $U \to \mathcal{Z}$). Hence $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact by another application of Properties of Stacks, Lemma 100.6.2

Assume (2). Let $Z \to \mathcal{Y}$ be a morphism, where $Z$ is a scheme. We have to show that the morphism of algebraic spaces $p : Z \times _\mathcal {Y} \mathcal{X} \to Z$ is quasi-compact. Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times _\mathcal {Y} \mathcal{Z}$ and the algebraic space $U \times _\mathcal {Y} \mathcal{Z}$ is quasi-compact by assumption (2). Hence $p$ is quasi-compact, see Morphisms of Spaces, Lemma 67.8.8.
$\square$

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