The Stacks project

Proof. By Morphisms of Spaces, Lemma 67.9.7 the morphism $f$ is quasi-compact. We just define $Y'$ as the normalization of $Y$ in $X$, so (5) and (2) hold automatically. By Morphisms of Spaces, Lemma 67.48.9 we see that (4) holds. The morphism $f'$ is universally closed by Morphisms of Spaces, Lemma 67.40.6. It is quasi-compact by Morphisms of Spaces, Lemma 67.8.9 and quasi-separated by Morphisms of Spaces, Lemma 67.4.10.

To show the remaining statements we may assume the base $Y$ is affine (as taking normalization commutes with étale localization). Say $Y = \mathop{\mathrm{Spec}}(R)$. Then $Y' = \mathop{\mathrm{Spec}}(A)$ with $A = \Gamma (X, \mathcal{O}_ X)$ an integral $R$-algebra. Thus it is clear that $f'_*\mathcal{O}_ X$ is $\mathcal{O}_{Y'}$ (because $f'_*\mathcal{O}_ X$ is quasi-coherent, by Morphisms of Spaces, Lemma 67.11.2, and hence equal to $\widetilde{A}$). This proves (3).

Let us show that $f'$ is surjective. As $f'$ is universally closed (see above) the image of $f'$ is a closed subset $V(I) \subset Y' = \mathop{\mathrm{Spec}}(A)$. Pick $h \in I$. Then $h|_ X = f^\sharp (h)$ is a global section of the structure sheaf of $X$ which vanishes at every point. As $X$ is quasi-compact this means that $h|_ X$ is a nilpotent section, i.e., $h^ n|X = 0$ for some $n > 0$. But $A = \Gamma (X, \mathcal{O}_ X)$, hence $h^ n = 0$. In other words $I$ is contained in the Jacobson radical of $A$ and we conclude that $V(I) = Y'$ as desired. $\square$