Lemma 98.4.3. Let

be a fibre product of algebraic stacks. Then the map of sets of points

is surjective.

Lemma 98.4.3. Let

\[ \xymatrix{ \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Z} \ar[r] & \mathcal{Y} } \]

be a fibre product of algebraic stacks. Then the map of sets of points

\[ |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \longrightarrow |\mathcal{Z}| \times _{|\mathcal{Y}|} |\mathcal{X}| \]

is surjective.

**Proof.**
Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to \mathcal{X}$, $\mathop{\mathrm{Spec}}(L) \to \mathcal{Z}$, then the assumption that they agree as elements of $|\mathcal{Y}|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to \mathcal{X} \to \mathcal{Y}$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to \mathcal{Z} \to \mathcal{Y}$ are $2$-isomorphic. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$.
$\square$

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