The Stacks project

Proof. To prove $\mathcal{F}$ is a sheaf, let's check conditions (1) and (2) of Topologies, Lemma 34.8.15. Condition (1) holds because formation of $X^{awn}$ commutes with open coverings, see Morphisms, Lemma 29.47.7 and its proof.

Let $\pi : Y \to X$ be a surjective proper morphism. We have to show that the equalizer of the two maps

is equal to $\Gamma (X^{awn}, \mathcal{O}_{X^{awn}})$. Let $f$ be an element of this equalizer. Then we consider the morphism

Since $Y^{awn} \to X$ is universally closed, the scheme theoretic image $Z$ of $f$ is a closed subscheme of $\mathbf{A}^1_ X$ proper over $X$ and $f : Y^{awn} \to Z$ is surjective. See Morphisms, Lemma 29.41.10. Thus $Z \to X$ is finite (Morphisms, Lemma 29.44.11) and surjective.

Let $k$ be a field and let $z_1, z_2 : \mathop{\mathrm{Spec}}(k) \to Z$ be two morphisms equalized by $Z \to X$. We claim that $z_1 = z_2$. It suffices to show the images $\lambda _ i = z_ i^*f \in k$ agree (as the structure sheaf of $Z$ is generated by $f$ over the structure sheaf of $X$). To see this we choose a field extension $K/k$ and morphisms $y_1, y_2 : \mathop{\mathrm{Spec}}(K) \to Y^{awn}$ such that $z_ i \circ (\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(k)) = f \circ y_ i$. This is possible by the surjectivity of the map $Y^{awn} \to Z$. Choose an algebraically closed extension $\Omega /k$ of very large cardinality. For any $k$-algebra maps $\sigma _ i : K \to \Omega $ we obtain

Since the canonical morphism $(Y \times _ X Y)^{awn} \to Y^{awn} \times _ X Y^{awn}$ is a universal homeomorphism and since $\Omega $ is algebraically closed, we can lift the composition above uniquely to a morphism $\mathop{\mathrm{Spec}}(\Omega ) \to (Y \times _ X Y)^{awn}$. Since $f$ is in the equalizer above, this proves that $\sigma _1(\lambda _1) = \sigma _2(\lambda _2)$. An easy lemma about field extensions shows that this implies $\lambda _1 = \lambda _2$; details omitted.

We conclude that $Z \to X$ is universally injective, i.e., $Z \to X$ is injective on points and induces purely inseparated residue field extensions (Morphisms, Lemma 29.10.2). All in all we conclude that $Z \to X$ is a universal homeomorphism, see Morphisms, Lemma 29.45.5.

Let $g : X^{awn} \to Z$ be the map obtained from the universal property of $X^{awn}$. Then $Y^{awn} \to X^{awn} \to Z$ and $f : Y^{awn} \to Z$ are two morphisms over $X$. By the universal property of $Y^{awn} \to Y$ the two corresponding morphisms $Y^{awn} \to Y \times _ X Z$ over $Y$ have to be equal. This implies that $g \circ \pi ^{wan} = f$ as morphisms into $\mathbf{A}^1_ X$ and we conclude that $g \in \Gamma (X^{awn}, \mathcal{O}_{X^{awn}})$ is the element we were looking for. $\square$