The Stacks project

Proof. Let $W \to Y$ be a surjective étale morphism where $W$ is a scheme. Then $W \times _ Y X$ is a scheme as well. By Lemma 67.48.4 the algebraic space $W \times _ Y Y'$ is representable and is the normalization of the scheme $W$ in the scheme $W \times _ Y X$. Picture

By More on Morphisms, Lemma 37.43.1 the result of the lemma holds over $W$. Let $V' \subset W \times _ Y Y'$ be the open subscheme such that

1. $(1, f')^{-1}(V') \to V'$ is an isomorphism, and

2. $(1, f')^{-1}(V') \subset W \times _ Y X$ is the set of points at which $(1, f)$ is quasi-finite.

By Lemma 67.34.7 there is a maximal open set of points $U \subset X$ where $f$ is quasi-finite and $W \times _ Y U = (1, f')^{-1}(V')$. The morphism $f'|_ U : U \to Y'$ is an open immersion by Lemma 67.12.1 as its base change to $W$ is the isomorphism $(1, f')^{-1}(V') \to V'$ followed by the open immersion $V' \to W \times _ Y Y'$. Setting $U' = \mathop{\mathrm{Im}}(U \to Y')$ finishes the proof (omitted: the verification that $(f')^{-1}(U') = U$). $\square$

Proof. By Lemma 67.51.1 the morphism $f$ is representable. Hence we may apply Lemma 67.52.1. Thus there exists an open subspace $U' \subset Y'$ such that $(f')^{-1}(U') = X$ (!) and $X \to U'$ is an isomorphism! In other words, $f'$ is an open immersion. Note that $f'$ is quasi-compact as $f$ is quasi-compact and $\nu : Y' \to Y$ is separated (Lemma 67.8.9). Hence for every affine scheme $Z$ and morphism $Z \to Y$ the fibre product $Z \times _ Y X$ is a quasi-compact open subscheme of the affine scheme $Z \times _ Y Y'$. Hence $f$ is quasi-affine by definition. $\square$