Lemma 101.37.6. Let \mathcal{Z} be an algebraic stack. Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks over \mathcal{Z}. If \mathcal{X} is universally closed over \mathcal{Z} and f is surjective then \mathcal{Y} is universally closed over \mathcal{Z}. In particular, if also \mathcal{Y} is separated and of finite type over \mathcal{Z}, then \mathcal{Y} is proper over \mathcal{Z}.
Proof. Assume \mathcal{X} is universally closed and f surjective. Denote p : \mathcal{X} \to \mathcal{Z}, q : \mathcal{Y} \to \mathcal{Z} the structure morphisms. Let \mathcal{Z}' \to \mathcal{Z} be a morphism of algebraic stacks. The base change f' : \mathcal{X}' \to \mathcal{Y}' of f by \mathcal{Z}' \to \mathcal{Z} is surjective (Properties of Stacks, Lemma 100.5.3) and the base change p' : \mathcal{X}' \to \mathcal{Z}' of p is closed. If T \subset |\mathcal{Y}'| is closed, then (f')^{-1}(T) \subset |\mathcal{X}'| is closed, hence p'((f')^{-1}(T)) = q'(T) is closed. So q' is closed. \square
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