The Stacks project

Proof. Let $T$ be an algebraic space over $S$. Let $\{ T_ i \to T\} $ be an fppf covering (by algebraic spaces). Let $s_ i = (a_ i, Z_ i)$ be pairs over $T_ i$ satisfying 79.12.0.1 such that we have $s_ i|_{T_ i \times _ T T_ j} = s_ j|_{T_ i \times _ T T_ j}$. First, this implies in particular that $a_ i$ and $a_ j$ define the same morphism $T_ i \times _ T T_ j \to Y$. By Descent on Spaces, Lemma 74.7.2 we deduce that there exists a unique morphism $a : T \to Y$ such that $a_ i$ equals the composition $T_ i \to T \to Y$. Second, this implies that $Z_ i \subset T_ i \times _ Y X$ are open subspaces whose inverse images in $(T_ i \times _ T T_ j) \times _ Y X$ are equal. Since $\{ T_ i \times _ Y X \to T \times _ Y X\} $ is an fppf covering we deduce that there exists a unique open subspace $Z \subset T \times _ Y X$ which restricts back to $Z_ i$ over $T_ i$, see Descent on Spaces, Lemma 74.7.1. We claim that the projection $Z \to T$ is finite. This follows as being finite is local for the fpqc topology, see Descent on Spaces, Lemma 74.11.23.

Note that the result of the preceding paragraph in particular implies (1).

Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say “pair” we mean a pair satisfying conditions 79.12.0.1.

Let $v : T \to (X/Y)_{fin}$ be a natural transformation. Choose a scheme $U$ and a surjective étale morphism $p : U \to T$. Then $v(p) \in (X/Y)_{fin}(U)$ corresponds to a pair $(a_ U, Z_ U)$ over $U$. Let $R = U \times _ T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_ U, Z_ U)$ by $s$ and $t$ agree. Hence, since $\{ U \to T\} $ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, Z)$ over $T$.

Conversely, let $(a, Z)$ be a pair over $T$. Let $U \to T$, $R = U \times _ T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, Z)|_ U$ gives rise to a transformation of functors $v : h_ U \to (X/Y)_{fin}$ by the Yoneda lemma (Categories, Lemma 4.3.5). As the two pullbacks $s^*(a, Z)|_ U$ and $t^*(a, Z)|_ U$ are equal, we see that $v$ coequalizes the two maps $h_ t, h_ s : h_ R \to h_ U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma 65.9.1 and since $(X/Y)_{fin}$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to (X/Y)_{fin}$.

We omit the verification that the two constructions above are mutually inverse. $\square$

Proof. If $(a, Z)$ is a pair over $T$ for $X'/Y$, then $(a, j(Z))$ is a pair over $T$ for $X/Y$. $\square$

Proof. Lemma 79.12.2 gives us an injective map $(X'/Y)_{fin} \to (X/Y)_{fin}$. Morphisms of Spaces, Lemma 67.34.7 assures us that formation of $X'$ commutes with base change. Hence everything comes down to proving that if $Z \subset X$ is an open subspace such that $f|_ Z : Z \to Y$ is finite, then $Z \subset X'$. This is true because a finite morphism is locally quasi-finite, see Morphisms of Spaces, Lemma 67.45.8. $\square$

Proof. A finite morphism of algebraic spaces is universally closed by Morphisms of Spaces, Lemma 67.45.9. Since $f$ is separated so is the morphism $T \times _ Y X \to T$, see Morphisms of Spaces, Lemma 67.4.4. Thus the closedness of $Z$ follows from Morphisms of Spaces, Lemma 67.40.6. $\square$

Proof. The second statement follows from the first as the absolute diagonal is the composition of the relative diagonal and a base change of the diagonal of $Y$ (which is representable by schemes), see Spaces, Section 65.3. To prove the first assertion we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 79.12.0.1 there is an open subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

Let us construct $V$. Note that $Z_1 \cap Z_2$ is open in $Z_1$ and in $Z_2$. Since $\text{pr}_0|_{Z_ i} : Z_ i \to T$ is finite, hence proper (see Morphisms of Spaces, Lemma 67.45.9) we see that

is closed in $T$. Now it is clear that $V = T \setminus E$ works. $\square$

Proof. This is formal from the description of the diagonal in Lemma 79.12.7 but we write it out since it is an important step in the development of the theory. We have to check that for any scheme $T$ over $S$ and a morphism $T \to (X/Y)_{fin}$ the projection map

is étale. Note that

Applying the result of Lemma 79.12.7 we see that $T \times _{(X/Y)_{fin}} U$ is represented by an open subscheme of $T \times _ Y U$. As the projection $T \times _ Y U \to T$ is étale by Morphisms of Spaces, Lemma 67.39.4 we conclude. $\square$

Proof. It follows immediately from the definitions that the sheaf $(X'/Y')_{fin}$ is equal to the sheaf $Y' \times _ Y (X/Y)_{fin}$. $\square$

Proof. Note that the assertion makes sense by the result of Lemma 79.12.7 on the diagonal of $(X/Y)_{fin}$, see Spaces, Lemma 65.5.10. Let $V$ be a scheme and let $V \to Y$ be a surjective étale morphism. By Lemma 79.12.9 the morphism $(V \times _ Y X/V)_{fin} \to (X/Y)_{fin}$ is a base change of the map $V \to Y$ and hence is surjective and étale, see Spaces, Lemma 65.5.5. Hence it suffices to prove the lemma for $(V \times _ Y X/V)_{fin}$. (Here we implicitly use that the composition of representable, surjective, and étale transformations of functors is again representable, surjective, and étale, see Spaces, Lemmas 65.3.2 and 65.5.4, and Morphisms, Lemmas 29.9.2 and 29.36.3.) Note that the properties of being separated and locally quasi-finite are preserved under base change, see Morphisms of Spaces, Lemmas 67.4.4 and 67.27.4. Hence $V \times _ Y X \to V$ is separated and locally quasi-finite as well, and by Morphisms of Spaces, Proposition 67.50.2 we see that $V \times _ Y X$ is a scheme as well. Thus we may assume that $f : X \to Y$ is a separated and locally quasi-finite morphism of schemes.

Pick a point $y \in Y$. Pick $x_1, \ldots , x_ n \in X$ points lying over $y$. Pick an étale neighbourhood $a : (U, u) \to (Y, y)$ and a decomposition

as in More on Morphisms, Lemma 37.41.5. Pick any subset

Given these choices we obtain a pair $(a, Z)$ with $Z = \bigcup _{(i, j) \in I} V_{i, j}$ which satisfies conditions 79.12.0.1. In other words we obtain a morphism $U \to (X/Y)_{fin}$. The construction of this morphism depends on all the things we picked above, so we should really write

This morphism is étale by Lemma 79.12.8.

Claim: The disjoint union of all of these is surjective onto $(X/Y)_{fin}$. It is clear that if the claim holds, then the lemma is true.

To show surjectivity we have to show the following (see Spaces, Remark 65.5.2): Given a scheme $T$ over $S$, a point $t \in T$, and a map $T \to (X/Y)_{fin}$ we can find a datum $(y, n, x_1, \ldots , x_ n, a, I)$ as above such that $t$ is in the image of the projection map

To prove this we may clearly replace $T$ by $\mathop{\mathrm{Spec}}(\overline{\kappa (t)})$ and $T \to (X/Y)_{fin}$ by the composition $\mathop{\mathrm{Spec}}(\overline{\kappa (t)}) \to T \to (X/Y)_{fin}$. In other words, we may assume that $T$ is the spectrum of an algebraically closed field.

Let $T = \mathop{\mathrm{Spec}}(k)$ be the spectrum of an algebraically closed field $k$. The morphism $T \to (X/Y)_{fin}$ is given by a pair $(T \to Y, Z)$ satisfying conditions 79.12.0.1. Here is a picture:

Let $y \in Y$ be the image point of $T \to Y$. Since $Z$ is finite over $k$ it has finitely many points. Thus there exist finitely many points $x_1, \ldots , x_ n \in X$ such that the image of $Z$ in $X$ is contained in $\{ x_1, \ldots , x_ n\} $. Choose $a : (U, u) \to (Y, y)$ adapted to $y$ and $x_1, \ldots , x_ n$ as above, which gives the diagram

Since $k$ is algebraically closed and $\kappa (y) \subset \kappa (u)$ is finite separable we may factor the morphism $T = \mathop{\mathrm{Spec}}(k) \to Y$ through the morphism $u = \mathop{\mathrm{Spec}}(\kappa (u)) \to \mathop{\mathrm{Spec}}(\kappa (y)) = y \subset Y$. With this choice we obtain the commutative diagram:

We know that the image of the left upper arrow ends up in $\coprod V_{i, j}$. Recall also that $Z$ is an open subscheme of $\mathop{\mathrm{Spec}}(k) \times _ Y X$ by definition of $(X/Y)_{fin}$ and that the right hand square is a fibre product square. Thus we see that

is an open subscheme. By construction (see More on Morphisms, Lemma 37.41.5) each $V_{i, j}$ has a unique point $v_{i, j}$ lying over $u$ with purely inseparable residue field extension $\kappa (v_{i, j})/\kappa (u)$. Hence each scheme $\mathop{\mathrm{Spec}}(k) \times _ U V_{i, j}$ has exactly one point. Thus we see that

for a unique subset $I \subset \{ (i, j) \mid 1 \leq i \leq n, \ 1 \leq j \leq m_ i\} $. Unwinding the definitions this shows that

with $I$ as found above is nonempty as desired. $\square$

Proof. By Lemma 79.12.3 we may replace $X$ by the open subscheme which is locally quasi-finite over $Y$. Hence we may assume that $f$ is separated and locally quasi-finite. We will check the three conditions of Spaces, Definition 65.6.1. Condition (1) follows from Lemma 79.12.1. Condition (2) follows from Lemma 79.12.7. Finally, condition (3) follows from Lemma 79.12.10. Thus $(X/Y)_{fin}$ is an algebraic space. Moreover, that lemma shows that there exists a commutative diagram

with horizontal arrow surjective and étale and south-east arrow étale. By Properties of Spaces, Lemma 66.16.3 this implies that the south-west arrow is étale as well. $\square$

Proof. Since $f$ is in particular separated and locally of finite type (see Morphisms of Spaces, Lemma 67.28.5) we see that $(X/Y)_{fin}$ is an algebraic space by Proposition 79.12.11. To prove that $(X/Y)_{fin} \to Y$ is separated we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 79.12.0.1 there is a closed subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

In the proof of Lemma 79.12.7 we have seen that $V = T' \setminus E$ is an open subscheme of $T'$ with closed complement

Thus everything comes down to showing that $E$ is also open. By Lemma 79.12.4 we see that $Z_1$ and $Z_2$ are closed in $T' \times _ Y X$. Hence $Z_1 \setminus Z_1 \cap Z_2$ is open in $Z_1$. As $f$ is flat and locally of finite presentation, so is $\text{pr}_0|_{Z_1}$. This is true as $Z_1$ is an open subspace of the base change $T' \times _ Y X$, and Morphisms of Spaces, Lemmas 67.28.3 and Lemmas 67.30.4. Hence $\text{pr}_0|_{Z_1}$ is open, see Morphisms of Spaces, Lemma 67.30.6. Thus $\text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right)$ is open and it follows that $E$ is open as desired.

We have already seen that $(X/Y)_{fin} \to Y$ is étale, see Proposition 79.12.11. Hence now we know it is locally quasi-finite (see Morphisms of Spaces, Lemma 67.39.5) and separated, hence representable by Morphisms of Spaces, Lemma 67.51.1. The final assertion is clear (if you like you can use Morphisms of Spaces, Proposition 67.50.2). $\square$

Proof. Omitted. Hint: Given a pair $(a, Z)$ over $T$ as in (79.12.0.1) the inverse image of $Z$ by $(1_ T, \sigma \circ a) : T \to T \times _ Y X$ is the open subscheme of $T$ we are looking for. $\square$