Proof.
Parts (1) and (2) are a consequence of (3) and (4) for S = B = \mathop{\mathrm{Spec}}(\mathbf{Z}) (see Properties of Spaces, Definition 66.3.1). Consider the commutative diagram
\xymatrix{ X \ar[d] \ar[rr]_{\Delta _{X/B}} & & X \times _ B X \ar[d] \\ Y \ar[rr]^{\Delta _{Y/B}} & & Y \times _ B Y }
The left vertical arrow is surjective (i.e., universally surjective). The right vertical arrow is universally closed as a composition of the universally closed morphisms X \times _ B X \to X \times _ B Y \to Y \times _ B Y. Hence it is also quasi-compact, see Lemma 67.9.7.
Assume X is quasi-separated over B, i.e., \Delta _{X/B} is quasi-compact. Then if Z is quasi-compact and Z \to Y \times _ B Y is a morphism, then Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y is surjective and Z \times _{Y \times _ B Y} X is quasi-compact by our remarks above. We conclude that \Delta _{Y/B} is quasi-compact, i.e., Y is quasi-separated over B.
Assume X is separated over B, i.e., \Delta _{X/B} is a closed immersion. Then if Z is affine, and Z \to Y \times _ B Y is a morphism, then Z \times _{Y \times _ B Y} X \to Z \times _{Y \times _ B Y} Y is surjective and Z \times _{Y \times _ B Y} X \to Z is universally closed by our remarks above. We conclude that \Delta _{Y/B} is universally closed. It follows that \Delta _{Y/B} is representable, locally of finite type, a monomorphism (see Lemma 67.4.1) and universally closed, hence a closed immersion, see Étale Morphisms, Lemma 41.7.2 (and also the abstract principle Spaces, Lemma 65.5.8). Thus Y is separated over B.
\square
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