**Proof.**
Assume (1). Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a morphism from the spectrum of a field into $X$. By assumption the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ X T \to \mathop{\mathrm{Spec}}(K)$ is surjective. Hence there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _ X T$ such that the left square in the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _ X T \ar[d] \ar[r] & T \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-x & X } \]

is commutative. This shows that $|f| : |T| \to |X|$ is surjective.

Assume (2). Let $Z \to X$ be a morphism where $Z$ is a scheme. We have to show that the morphism of schemes $Z \times _ X T \to T$ is surjective, i.e., that $|Z \times _ X T| \to |Z|$ is surjective. This follows from (2) and Lemma 65.4.3.
$\square$

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