The Stacks project

Proof. We are going to use the fact that $\Delta _{X/S}$ is representable (by definition of an algebraic space) and that it satisfies properties (2) – (5), see Spaces, Lemma 65.13.1. Note that we have a factorization

of the diagonal $\Delta _{X/S} : X \to X \times _ S X$. Since $X \times _ Y X \to X \times _ S X$ is a monomorphism, and since $\Delta _{X/S}$ is representable, it follows formally that $\Delta _{X/Y}$ is representable. In particular, the rest of the statements now make sense, see Section 67.3.

Choose a surjective étale morphism $U \to X$, with $U$ a scheme. Consider the diagram

Both squares are cartesian, hence so is the outer rectangle. The top row consists of schemes, and the vertical arrows are surjective étale morphisms. By Spaces, Lemma 65.11.4 the properties (2) – (5) for $\Delta _{X/Y}$ are equivalent to those of $R \to U \times _ Y U$. In the proof of Spaces, Lemma 65.13.1 we have seen that $R \to U \times _ S U$ has properties (2) – (5). The morphism $U \times _ Y U \to U \times _ S U$ is a monomorphism of schemes. These facts imply that $R \to U \times _ Y U$ have properties (2) – (5).

Namely: For (3), note that $R \to U \times _ Y U$ is a monomorphism as the composition $R \to U \times _ S U$ is a monomorphism. For (2), note that $R \to U \times _ Y U$ is locally of finite type, as the composition $R \to U \times _ S U$ is locally of finite type (Morphisms, Lemma 29.15.8). A monomorphism which is locally of finite type is locally quasi-finite because it has finite fibres (Morphisms, Lemma 29.20.7), hence (5). A monomorphism is separated (Schemes, Lemma 26.23.3), hence (4). $\square$

Proof. This is true, via the general principle Spaces, Lemma 65.5.8, because a closed immersion of schemes is an immersion and is quasi-compact. $\square$

Proof. Let $f : X \to Y$ and $Y' \to Y$ be morphisms of algebraic spaces. Let $f' : X' \to Y'$ be the base change of $f$ by $Y' \to Y$. Then $\Delta _{X'/Y'}$ is the base change of $\Delta _{X/Y}$ by the morphism $X' \times _{Y'} X' \to X \times _ Y X$. By the results of Section 67.3 each of the properties of the diagonal used in Definition 67.4.2 is stable under base change. Hence the lemma is true. $\square$

Proof. By general category theory the following diagram

is a fibre product diagram. Hence $i$ is the base change of the diagonal morphism $\Delta _{Z/T}$. Thus the lemma follows from Lemma 67.4.1, and the material in Section 67.3. $\square$

Proof. This is a special case of Lemma 67.4.5 applied to the morphism $X = X \times _ Y Y \to X \times _ T Y$. $\square$

Proof. This is a special case of Lemma 67.4.6 applied to $g = s$ so the morphism $i = s : T \to T \times _ T X$. $\square$

Proof. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces to which the axiom in question applies. The diagonal $\Delta _{X/Z}$ is the composition

Our separation axiom is defined by requiring the diagonal to have some property $\mathcal{P}$. By Lemma 67.4.5 above we see that the second arrow also has this property. Hence the lemma follows since the composition of (representable) morphisms with property $\mathcal{P}$ also is a morphism with property $\mathcal{P}$, see Section 67.3. $\square$

Proof. Parts (4), (5), and (6) follow immediately from Lemma 67.4.8 and Spaces, Definition 65.13.2. Parts (1), (2), and (3) reduce to parts (4), (5), and (6) by thinking of $X$ and $Y$ as algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Properties of Spaces, Definition 66.3.1. $\square$

Proof. Consider the factorization

of the diagonal morphism of $g \circ f$. In any case the last morphism is a monomorphism. Hence for any scheme $T$ and morphism $T \to X \times _ Y X$ we have the equality

Hence the result is clear. $\square$

Proof. Parts (4), (5), and (6) follow immediately from Lemma 67.4.10 and Spaces, Definition 65.13.2. Parts (1), (2), and (3) follow from parts (4), (5), and (6) by thinking of $X$ and $Y$ as algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Properties of Spaces, Definition 66.3.1. $\square$

Proof. We will repeatedly use Lemma 67.4.4 without further mention. In particular, it is clear that (1) implies (2) and (2) implies (3).

Let us prove that (3) and (4) are equivalent. Note that if $Z$ is an affine scheme, then the morphism $Z \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is a separated morphism as a morphism of algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. If $Z \times _ Y X \to Z$ is $\mathcal{P}$, then $Z \times _ Y X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is $\mathcal{P}$ as a composition (see Lemma 67.4.8). Hence the algebraic space $Z \times _ Y X$ is $\mathcal{P}$. Conversely, if the algebraic space $Z \times _ Y X$ is $\mathcal{P}$, then $Z \times _ Y X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is $\mathcal{P}$, and hence by Lemma 67.4.10 we see that $Z \times _ Y X \to Z$ is $\mathcal{P}$.

Let us prove that (3) implies (5). Assume (3). Let $V$ be a scheme and let $V \to Y$ be étale surjective. We have to show that $V \times _ Y X \to V$ has property $\mathcal{P}$. In other words, we have to show that the morphism

has the corresponding property (i.e., is a closed immersion, immersion, or quasi-compact). Let $V = \bigcup V_ j$ be an affine open covering of $V$. By assumption we know that each of the morphisms

does have the corresponding property. Since being a closed immersion, immersion, quasi-compact immersion, or quasi-compact is Zariski local on the target, and since the $V_ j$ cover $V$ we get the desired conclusion.

Let us prove that (5) implies (1). Let $V \to Y$ be as in (5). Then we have the fibre product diagram

By assumption the left vertical arrow is a closed immersion, immersion, quasi-compact immersion, or quasi-compact. It follows from Spaces, Lemma 65.5.6 that also the right vertical arrow is a closed immersion, immersion, quasi-compact immersion, or quasi-compact.

It is clear that (1) implies (6) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (6). Choose schemes $V_ i$ and surjective étale morphisms $V_ i \to Y_ i$. Note that the morphisms $V_ i \times _ Y X \to V_ i$ have $\mathcal{P}$ as they are base changes of the morphisms $f^{-1}(Y_ i) \to Y_ i$. Set $V = \coprod V_ i$. Then $V \to Y$ is a morphism as in (5) (details omitted). Hence (6) implies (5) and we are done. $\square$

Proof. This is the equivalence of (1) and (2) of Lemma 67.4.12 combined with the fact that any morphism of schemes is locally separated, see Schemes, Lemma 26.21.2. $\square$