The Stacks project

Proof. We will prove (1) and omit the proof of (2); also the final assertion will follow from the construction of the factorization. We will use Lemma 29.45.5 without further mention. First, let $Y \to X^{Y/n} \to X$ be the normalization of $X$ in $Y$, see Definition 29.53.3. For $Y \to X' \to X$ as in (1), we obtain a unique morphism $X^{Y/n} \to X'$ compatible with the given morphisms, see Lemma 29.53.4. Thus it suffices to prove the lemma with $f$ replaced by $X^{Y/n} \to X$. This reduces us to the case studied in the next paragraph.

Assume $f$ is integral (the rest of the proof works more generally if $f$ is affine). Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine open of $X$ and let $V = f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$ be the inverse image in $Y$. Then $A \to B$ is a ring map which induces a dominant morphism on spectra. By Lemma 29.55.1 we obtain an $A$-subalgebra $B' \subset B$ such that setting $U^{V/wn} = \mathop{\mathrm{Spec}}(B')$ the factorization $V \to U^{V/wn} \to U$ is initial in the category of factorizations $V \to U' \to U$ where $U' \to U$ is a universal homeomorphism.

If $U_1 \subset U_2 \subset X$ are affine opens, then setting $V_ i = f^{-1}(U_ i)$ we obtain a canonical morphism

over $U_1$ by the universal property of $U_1^{V_1/wn}$. These morphisms satisfy a natural functoriality which we leave to the reader to formulate and prove. Furthermore, the morphism $\rho _{U_1}^{U_2}$ is an isomorphism; this follows from Lemma 29.55.2 provided that $U_1 \subset U_2$ is a standard open and in the general case can be reduced to this case by the functorial nature of these maps and Schemes, Lemma 26.11.5 (details omitted). Thus by relative glueing (Constructions, Lemma 27.2.1) we obtain a morphism $X^{Y/wn} \to X$ which restricts to $U^{V/wn} \to U$ over $U$ compatibly with the $\rho _{U_1}^{U_2}$. Of course, the morphisms $V \to U^{V/wn}$ glue to a morphism $Y \to X^{Y/wn}$ (see Constructions, Remark 27.2.3) and we get our factorization $Y \to X^{Y/wn} \to X$ where the second morphism is a universal homeomorphism.

Finally, let $Y \to X' \to X$ be a factorization as in (1). With $V \to U^{V/wn} \to U \subset X$ as above, we obtain a factorization $V \to U \times _ X X' \to U$ where the second arrow is a universal homeomorphism and we obtain a unique morphism $g_ U : U^{V/wn} \to U \times _ X X'$ over $U$ by the universal property of $U^{V/wn}$. These $g_ U$ are compatible with the morphisms $\rho _{U_1}^{U_2}$; details omitted. Hence there is a unique morphism $g : X^{Y/wn} \to X'$ over $X$ agreeing with $g_ U$ over $U$, see Constructions, Remark 27.2.3. This proves that $Y \to X^{Y/wn} \to X$ is initial in our category and the proof is complete. $\square$