Proof.
Assume (1). Let T \to \mathcal{Y} be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces T \times _\mathcal {Y} \mathcal{X} \to T is surjective. By Morphisms of Spaces, Definition 67.5.2 this means we have to show that |T \times _\mathcal {Y} \mathcal{X}| \to |T| is surjective. Applying Lemma 100.4.3 we see that this follows from (1).
Conversely, assume (2). Let y : \mathop{\mathrm{Spec}}(K) \to \mathcal{Y} be a morphism from the spectrum of a field into \mathcal{Y}. By assumption the morphism \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \to \mathop{\mathrm{Spec}}(K) of algebraic spaces is surjective. By Morphisms of Spaces, Definition 67.5.2 this means there exists a field extension K'/K and a morphism \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} such that the left square of the diagram
\xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-y & \mathcal{Y} }
is commutative. This shows that |X| \to |\mathcal{Y}| is surjective.
\square
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