The Stacks project

Proof. Consider the category $\mathcal{C}$ whose objects are the schemes $U$ over $S$ such that every quasi-compact open of $U$ has finitely many irreducible components and whose morphisms are those morphisms $g : U \to V$ of schemes over $S$ such that every generic point of an irreducible component of $U$ is mapped to the generic point of an irreducible component of $V$. We have already shown that

1. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have a normalization morphism $\nu _ U : U^\nu \to U$ as in Morphisms, Definition 29.54.1,

2. for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the morphism $\nu _ U$ is integral and $U^\nu$ is a normal scheme, see Morphisms, Lemma 29.54.5,

3. for every $g : U \to V \in \text{Arrows}(\mathcal{C})$ there is a unique morphism $g^\nu : U^\nu \to V^\nu$ such that $\nu _ V \circ g^\nu = g \circ \nu _ U$, see Morphisms, Lemma 29.54.5 part (4) applied to the composition $X^\nu \to X \to Y$,

4. if $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $g : U \to V$ is étale or smooth, then $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $g \in \text{Arrows}(\mathcal{C})$ and the morphism $g^\nu$ induces an isomorphism $U^\nu \to U \times _ V V^\nu$, see Lemma 67.49.4 and More on Morphisms, Lemma 37.19.3.

Our task is to extend this construction to the corresponding category of algebraic spaces $X$ over $S$.

Let $X$ be an algebraic space over $S$ satisfying the equivalent conditions of Lemma 67.49.1. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Set $R = U \times _ X U$ with projections $s, t : R \to U$ and $j = (t, s) : R \to U \times _ S U$ so that $X = U/R$, see Spaces, Lemma 65.9.1. Observe that $U$ and $R$ are objects of $\mathcal{C}$ by our assumptions on $X$ and that the morphisms $s$ and $t$ are étale morphisms of schemes over $S$. By (a) we have the normalization morphisms $\nu _ U : U^\nu \to U$ and $\nu _ R : R^\nu \to R$, by (d) we have morphisms $s^\nu : R^\nu \to U^\nu$, $t^\nu : R^\nu \to U^\nu$ which define isomorphisms $R^\nu \to R \times _{s, U} U^\nu$ and $R^\nu \to U^\nu \times _{U, t} R$. It follows that $s^\nu$ and $t^\nu$ are étale (as they are isomorphic to base changes of étale morphisms). The induced morphism $j^\nu = (t^\nu , s^\nu ) : R^\nu \to U^\nu \times _ S U^\nu$ is a monomorphism as it is equal to the composition

The first arrow is the diagonal morphism of $\nu _ R$. (This tells us that $R^\nu$ is a subscheme of the restriction of $R$ to $U^\nu$.) A formal computation with fibre products using property (d) shows that $R^\nu \times _{s^\nu , U^\nu , t^\nu } R^\nu$ is the normalization of $R \times _{s, U, t} R$. Hence the étale morphism $c : R \times _{s, U, t} R \to R$ extends uniquely to $c^\nu$ by (d). The morphism $c^\nu$ is compatible with the projection $\text{pr}_{13} : U^\nu \times _ S U^\nu \times _ S U^\nu \to U^\nu \times _ S U^\nu$. Similarly, there are morphisms $i^\nu : R^\nu \to R^\nu$ compatible with the morphism $U^\nu \times _ S U^\nu \to U^\nu \times _ S U^\nu$ which switches factors and there is a morphism $e^\nu : U^\nu \to R^\nu$ compatible with the diagonal morphism $U^\nu \to U^\nu \times _ S U^\nu$. All in all it follows that $j^\nu : R^\nu \to U^\nu \times _ S U^\nu$ is an étale equivalence relation. At this point we may and do set $X^\nu = U^\nu /R^\nu$ (Spaces, Theorem 65.10.5). Then we see that we have $U^\nu = X^\nu \times _ X U$ by Groupoids, Lemma 39.20.7.

What have we shown in the previous paragraph is this: for every algebraic space $X$ over $S$ satisfying the equivalent conditions of Lemma 67.49.1 if we choose a surjective étale morphism $g : U \to X$ where $U$ is a scheme, then we obtain a cartesian diagram

of algebraic spaces. This immediately implies that $X^\nu$ is a normal algebraic space and that $\nu _ X$ is a integral morphism. This gives part (1) of the lemma.

We will show below that the morphism $\nu _ X : X^\nu \to X$ up to unique isomorphism is independent of the choice of $g$, but for now, if $X$ is a scheme, we choose $\text{id} : X \to X$ so that it is clear that we have part (2) of the lemma.

We still have to prove parts (3) and (4). Let $g : U \to X$ and $\nu _ X : X^\nu \to X$ and $g^\nu : U^\nu \to X^\nu$ be as above. Let $Z$ be a normal scheme and let $h : Z \to U$ and $a : Z \to X^\nu$ be morphisms over $S$ such that $g \circ h = \nu _ X \circ a$ and such that every irreducible component of $Z$ dominates an irreducible component of $U$ (via $h$). By Morphisms, Lemma 29.54.5 part (4) we obtain a unique morphism $h^\nu : Z \to U^\nu$ such that $h = \nu _ U \circ h^\nu$. Picture:

Observe that $a = g^\nu \circ h^\nu$. Namely, since the square with corners $X^\nu$, $X$, $U^\nu$, $U$ is cartesian, this follows immediately from the fact that $h^\nu$ is unique (given $h$). In other words, given $h : Z \to U$ as above (and not $a$) there is a unique morphism $a : Z \to X^\nu$ with $\nu _ X \circ a = g \circ h$.

Let $f : X \to Y$ be as in part (3) of the statement of the lemma. Suppose we have chosen surjective étale morphisms $U \to X$ and $V \to Y$ where $U$ and $V$ are schemes such that $f$ lifts to a morphism $g : U \to V$. Then $g \in \text{Arrows}(\mathcal{C})$ and we obtain a unique morphism $g^\nu : U^\nu \to V^\nu$ compatible with $\nu _ U$ and $\nu _ V$. However, then the two morphisms

must be the same by our comments in the previous paragraph (applied with $Y$ in stead of $X$). Since $X^\nu$ is constructed by taking the quotient of $U^\nu$ by $R^\nu$ it follows that we obtain a (unique) morphism $f^\nu : X^\nu \to Y^\nu$ as stated in (3).

To see that the construction of $X^\nu$ is independent of the choice of $g : U \to X$ surjective étale, apply the construction in the previous paragraph to $\text{id} : X \to X$ and a morphism $U' \to U$ between étale coverings of $X$. This is enough because given any two étale coverings of $X$ there is a third one which dominates both. The reader shows that the morphism between the two normalizations constructed using either $U' \to X$ or $U \to X$ becomes an isomorphism after base change to $U'$ and hence was an isomorphism. We omit the details.

We omit the proof of (4) which is similar; hint use part (d) above. $\square$