Proof.
Assume (1), and let \mathcal{Z} \to \mathcal{Y} be as in (2). Choose a scheme V and a surjective smooth morphism V \to \mathcal{Z}. By assumption the morphism V \times _\mathcal {Y} \mathcal{X} \to V of algebraic spaces is universally submersive, in particular the map |V \times _\mathcal {Y} \mathcal{X}| \to |V| is submersive. By Properties of Stacks, Section 100.4 in the commutative diagram
\xymatrix{ |V \times _\mathcal {Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }
the horizontal arrows are open and surjective, and moreover
|V \times _\mathcal {Y} \mathcal{X}| \longrightarrow |V| \times _{|\mathcal{Z}|} |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}|
is surjective. Hence as the left vertical arrow is submersive it follows that the right vertical arrow is submersive. This proves (2). The implication (2) \Rightarrow (1) follows from the definitions.
\square
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