**Proof.**
Assume (2). Let $k$ be a field and let $y$ be an object of $\mathcal{Y}$ over $k$. Let $X_ y$ denote an algebraic space over $k$ representing the $2$-fibre product

\[ (\mathit{Sch}/\mathop{\mathrm{Spec}}(k))_{fppf} \times _{y, \mathcal{Y}, p} \mathcal{X}. \]

As we've assumed that $p$ is surjective we see that $X_ y$ is not empty. Hence we can find a field extension $K/k$ and a $K$-valued point $x$ of $X_ y$. Via the $2$-Yoneda lemma this corresponds to an object $x$ of $\mathcal{X}$ over $K$ together with an isomorphism $p(x) \cong y|_{\mathop{\mathrm{Spec}}(K)}$ and we see that (1) holds.

Assume (1). Choose a scheme $T$ and a $1$-morphism $y : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$. Let $X_ y$ be an algebraic space over $T$ representing the $2$-fibre product $(\mathit{Sch}/T)_{fppf} \times _{y, \mathcal{Y}, p} \mathcal{X}$. We have to show that $X_ y \to T$ is surjective. By Morphisms of Spaces, Definition 66.5.2 we have to show that $|X_ y| \to |T|$ is surjective. This means exactly that given a field $k$ over $T$ and a morphism $t : \mathop{\mathrm{Spec}}(k) \to T$ there exists a field extension $K/k$ and a morphism $x : \mathop{\mathrm{Spec}}(K) \to X_ y$ such that

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d] \ar[r]_ x & X_ y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ t & T } \]

commutes. By the $2$-Yoneda lemma this means exactly that we have to find $k \subset K$ and an object $x$ of $\mathcal{X}$ over $K$ such that $p(x) \cong t^*y|_{\mathop{\mathrm{Spec}}(K)}$. Hence (1) guarantees that this is the case and we win.
$\square$

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