## 75.7 Being proper over a base

This section is the analogue of Cohomology of Schemes, Section 30.26. As usual with material having to do with topology on the sets of points, we have to be careful translating the material to algebraic spaces.

Lemma 75.7.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $T \subset |X|$ be a closed subset. The following are equivalent

the morphism $Z \to Y$ is proper if $Z$ is the reduced induced algebraic space structure on $T$ (Properties of Spaces, Definition 66.12.5),

for some closed subspace $Z \subset X$ with $|Z| = T$ the morphism $Z \to Y$ is proper, and

for any closed subspace $Z \subset X$ with $|Z| = T$ the morphism $Z \to Y$ is proper.

**Proof.**
The implications (3) $\Rightarrow $ (1) and (1) $\Rightarrow $ (2) are immediate. Thus it suffices to prove that (2) implies (3). We urge the reader to find their own proof of this fact. Let $Z'$ and $Z''$ be closed subspaces with $T = |Z'| = |Z''|$ such that $Z' \to Y$ is a proper morphism of algebraic spaces. We have to show that $Z'' \to Y$ is proper too. Let $Z''' = Z' \cup Z''$ be the scheme theoretic union, see Morphisms of Spaces, Definition 67.14.4. Then $Z'''$ is another closed subspace with $|Z'''| = T$. This follows for example from the description of scheme theoretic unions in Morphisms of Spaces, Lemma 67.14.6. Since $Z'' \to Z'''$ is a closed immersion it suffices to prove that $Z''' \to Y$ is proper (see Morphisms of Spaces, Lemmas 67.40.5 and 67.40.4). The morphism $Z' \to Z'''$ is a bijective closed immersion and in particular surjective and universally closed. Then the fact that $Z' \to Y$ is separated implies that $Z''' \to Y$ is separated, see Morphisms of Spaces, Lemma 67.9.8. Moreover $Z''' \to Y$ is locally of finite type as $X \to Y$ is locally of finite type (Morphisms of Spaces, Lemmas 67.23.7 and 67.23.2). Since $Z' \to Y$ is quasi-compact and $Z' \to Z'''$ is a universal homeomorphism we see that $Z''' \to Y$ is quasi-compact. Finally, since $Z' \to Y$ is universally closed, we see that the same thing is true for $Z''' \to Y$ by Morphisms of Spaces, Lemma 67.40.7. This finishes the proof.
$\square$

Definition 75.7.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $T \subset |X|$ be a closed subset. We say *$T$ is proper over $Y$* if the equivalent conditions of Lemma 75.7.1 are satisfied.

The lemma used in the definition above is false if the morphism $f : X \to Y$ is not locally of finite type. Therefore we urge the reader not to use this terminology if $f$ is not locally of finite type.

Lemma 75.7.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $T' \subset T \subset |X|$ be closed subsets. If $T$ is proper over $Y$, then the same is true for $T'$.

**Proof.**
Omitted.
$\square$

Lemma 75.7.4. Let $S$ be a scheme. Consider a cartesian diagram of algebraic spaces over $S$

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

with $f$ locally of finite type. If $T$ is a closed subset of $|X|$ proper over $Y$, then $|g'|^{-1}(T)$ is a closed subset of $|X'|$ proper over $Y'$.

**Proof.**
Observe that the statement makes sense as $f'$ is locally of finite type by Morphisms of Spaces, Lemma 67.23.3. Let $Z \subset X$ be the reduced induced closed subspace structure on $T$. Denote $Z' = (g')^{-1}(Z)$ the scheme theoretic inverse image. Then $Z' = X' \times _ X Z = (Y' \times _ Y X) \times _ X Z = Y' \times _ Y Z$ is proper over $Y'$ as a base change of $Z$ over $Y$ (Morphisms of Spaces, Lemma 67.40.3). On the other hand, we have $T' = |Z'|$. Hence the lemma holds.
$\square$

Lemma 75.7.5. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces which are locally of finite type over $B$.

If $Y$ is separated over $B$ and $T \subset |X|$ is a closed subset proper over $B$, then $|f|(T)$ is a closed subset of $|Y|$ proper over $B$.

If $f$ is universally closed and $T \subset |X|$ is a closed subset proper over $B$, then $|f|(T)$ is a closed subset of $Y$ proper over $B$.

If $f$ is proper and $T \subset |Y|$ is a closed subset proper over $B$, then $|f|^{-1}(T)$ is a closed subset of $|X|$ proper over $B$.

**Proof.**
Proof of (1). Assume $Y$ is separated over $B$ and $T \subset |X|$ is a closed subset proper over $B$. Let $Z$ be the reduced induced closed subspace structure on $T$ and apply Morphisms of Spaces, Lemma 67.40.8 to $Z \to Y$ over $B$ to conclude.

Proof of (2). Assume $f$ is universally closed and $T \subset |X|$ is a closed subset proper over $B$. Let $Z$ be the reduced induced closed subspace structure on $T$ and let $Z'$ be the reduced induced closed subspace structure on $|f|(T)$. We obtain an induced morphism $Z \to Z'$. Denote $Z'' = f^{-1}(Z')$ the scheme theoretic inverse image. Then $Z'' \to Z'$ is universally closed as a base change of $f$ (Morphisms of Spaces, Lemma 67.40.3). Hence $Z \to Z'$ is universally closed as a composition of the closed immersion $Z \to Z''$ and $Z'' \to Z'$ (Morphisms of Spaces, Lemmas 67.40.5 and 67.40.4). We conclude that $Z' \to B$ is separated by Morphisms of Spaces, Lemma 67.9.8. Since $Z \to B$ is quasi-compact and $Z \to Z'$ is surjective we see that $Z' \to B$ is quasi-compact. Since $Z' \to B$ is the composition of $Z' \to Y$ and $Y \to B$ we see that $Z' \to B$ is locally of finite type (Morphisms of Spaces, Lemmas 67.23.7 and 67.23.2). Finally, since $Z \to B$ is universally closed, we see that the same thing is true for $Z' \to B$ by Morphisms of Spaces, Lemma 67.40.7. This finishes the proof.

Proof of (3). Assume $f$ is proper and $T \subset |Y|$ is a closed subset proper over $B$. Let $Z$ be the reduced induced closed subspace structure on $T$. Denote $Z' = f^{-1}(Z)$ the scheme theoretic inverse image. Then $Z' \to Z$ is proper as a base change of $f$ (Morphisms of Spaces, Lemma 67.40.3). Whence $Z' \to B$ is proper as the composition of $Z' \to Z$ and $Z \to B$ (Morphisms of Spaces, Lemma 67.40.4). This finishes the proof.
$\square$

Lemma 75.7.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $T_ i \subset |X|$, $i = 1, \ldots , n$ be closed subsets. If $T_ i$, $i = 1, \ldots , n$ are proper over $Y$, then the same is true for $T_1 \cup \ldots \cup T_ n$.

**Proof.**
Let $Z_ i$ be the reduced induced closed subscheme structure on $T_ i$. The morphism

\[ Z_1 \amalg \ldots \amalg Z_ n \longrightarrow X \]

is finite by Morphisms of Spaces, Lemmas 67.45.10 and 67.45.11. As finite morphisms are universally closed (Morphisms of Spaces, Lemma 67.45.9) and since $Z_1 \amalg \ldots \amalg Z_ n$ is proper over $S$ we conclude by Lemma 75.7.5 part (2) that the image $Z_1 \cup \ldots \cup Z_ n$ is proper over $S$.
$\square$

Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ X$-module. Then the support $\text{Supp}(\mathcal{F})$ of $\mathcal{F}$ is a closed subset of $|X|$, see Morphisms of Spaces, Lemma 67.15.2. Hence it makes sense to say “the support of $\mathcal{F}$ is proper over $Y$”.

Lemma 75.7.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ X$-module. The following are equivalent

the support of $\mathcal{F}$ is proper over $Y$,

the scheme theoretic support of $\mathcal{F}$ (Morphisms of Spaces, Definition 67.15.4) is proper over $Y$, and

there exists a closed subspace $Z \subset X$ and a finite type, quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$ such that (a) $Z \to Y$ is proper, and (b) $(Z \to X)_*\mathcal{G} = \mathcal{F}$.

**Proof.**
The support $\text{Supp}(\mathcal{F})$ of $\mathcal{F}$ is a closed subset of $|X|$, see Morphisms of Spaces, Lemma 67.15.2. Hence we can apply Definition 75.7.2. Since the scheme theoretic support of $\mathcal{F}$ is a closed subspace whose underlying closed subset is $\text{Supp}(\mathcal{F})$ we see that (1) and (2) are equivalent by Definition 75.7.2. It is clear that (2) implies (3). Conversely, if (3) is true, then $\text{Supp}(\mathcal{F}) \subset |Z|$ and hence $\text{Supp}(\mathcal{F})$ is proper over $Y$ for example by Lemma 75.7.3.
$\square$

Lemma 75.7.8. Let $S$ be a scheme. Consider a cartesian diagram of algebraic spaces over $S$

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

with $f$ locally of finite type. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ X$-module. If the support of $\mathcal{F}$ is proper over $Y$, then the support of $(g')^*\mathcal{F}$ is proper over $Y'$.

**Proof.**
Observe that the statement makes sense because $(g')*\mathcal{F}$ is of finite type by Modules on Sites, Lemma 18.23.4. We have $\text{Supp}((g')^*\mathcal{F}) = |g'|^{-1}(\text{Supp}(\mathcal{F}))$ by Morphisms of Spaces, Lemma 67.15.2. Thus the lemma follows from Lemma 75.7.4.
$\square$

Lemma 75.7.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $\mathcal{F}$, $\mathcal{G}$ be finite type, quasi-coherent $\mathcal{O}_ X$-module.

If the supports of $\mathcal{F}$, $\mathcal{G}$ are proper over $Y$, then the same is true for $\mathcal{F} \oplus \mathcal{G}$, for any extension of $\mathcal{G}$ by $\mathcal{F}$, for $\mathop{\mathrm{Im}}(u)$ and $\mathop{\mathrm{Coker}}(u)$ given any $\mathcal{O}_ X$-module map $u : \mathcal{F} \to \mathcal{G}$, and for any quasi-coherent quotient of $\mathcal{F}$ or $\mathcal{G}$.

If $Y$ is locally Noetherian, then the category of coherent $\mathcal{O}_ X$-modules with support proper over $Y$ is a Serre subcategory (Homology, Definition 12.10.1) of the abelian category of coherent $\mathcal{O}_ X$-modules.

**Proof.**
Proof of (1). Let $T$, $T'$ be the support of $\mathcal{F}$ and $\mathcal{G}$. Then all the sheaves mentioned in (1) have support contained in $T \cup T'$. Thus the assertion itself is clear from Lemmas 75.7.3 and 75.7.6 provided we check that these sheaves are finite type and quasi-coherent. For quasi-coherence we refer the reader to Properties of Spaces, Section 66.29. For “finite type” we refer the reader to Properties of Spaces, Section 66.30.

Proof of (2). The proof is the same as the proof of (1). Note that the assertions make sense as $X$ is locally Noetherian by Morphisms of Spaces, Lemma 67.23.5 and by the description of the category of coherent modules in Cohomology of Spaces, Section 69.12.
$\square$

Lemma 75.7.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type and $Y$ locally Noetherian. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module with support proper over $Y$. Then $R^ pf_*\mathcal{F}$ is a coherent $\mathcal{O}_ Y$-module for all $p \geq 0$.

**Proof.**
By Lemma 75.7.7 there exists a closed immersion $i : Z \to X$ with $g = f \circ i : Z \to Y$ proper and $\mathcal{F} = i_*\mathcal{G}$ for some coherent module $\mathcal{G}$ on $Z$. We see that $R^ pg_*\mathcal{G}$ is coherent on $S$ by Cohomology of Spaces, Lemma 69.20.2. On the other hand, $R^ qi_*\mathcal{G} = 0$ for $q > 0$ (Cohomology of Spaces, Lemma 69.12.9). By Cohomology on Sites, Lemma 21.14.7 we get $R^ pf_*\mathcal{F} = R^ pg_*\mathcal{G}$ and the lemma follows.
$\square$

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