The Stacks project

Proof. There is a canonical map $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Choose $0 \in I$. If $W_0 \subset X_0$ is an open subspace, then we have $f_0^{-1}W_0 = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}W_0$, see Lemma 70.4.1. Hence, if we can prove the lemma for inverse systems where $X_0$ is quasi-compact, then the lemma follows in general. Thus we may and do assume $X_0$ is quasi-compact.

Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = X_ i \times _{X_0} U_0$ and $U = X \times _{X_0} U_0$. Set $R_ i = U_ i \times _{X_ i} U_ i$ and $R = U \times _ X U$. Recall that $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and $R = \mathop{\mathrm{lim}}\nolimits R_ i$, see proof of Lemma 70.4.1. Recall that $|X| = |U|/|R|$ and $|X_ i| = |U_ i|/|R_ i|$. By Limits, Lemma 32.4.6 we have $|U| = \mathop{\mathrm{lim}}\nolimits |U_ i|$ and $|R| = \mathop{\mathrm{lim}}\nolimits |R_ i|$.

Surjectivity of $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Let $(x_ i) \in \mathop{\mathrm{lim}}\nolimits |X_ i|$. Denote $S_ i \subset |U_ i|$ the inverse image of $x_ i$. This is a finite nonempty set by the definition of decent spaces (Decent Spaces, Definition 68.6.1). Hence $\mathop{\mathrm{lim}}\nolimits S_ i$ is nonempty, see Categories, Lemma 4.21.7. Let $(u_ i) \in \mathop{\mathrm{lim}}\nolimits S_ i \subset \mathop{\mathrm{lim}}\nolimits |U_ i|$. By the above this determines a point $u \in |U|$ which maps to an $x \in |X|$ mapping to the given element $(x_ i)$ of $\mathop{\mathrm{lim}}\nolimits |X_ i|$.

Injectivity of $|X| \to \mathop{\mathrm{lim}}\nolimits |X_ i|$. Suppose that $x, x' \in |X|$ map to the same point of $\mathop{\mathrm{lim}}\nolimits |X_ i|$. Choose lifts $u, u' \in |U|$ and denote $u_ i, u'_ i \in |U_ i|$ the images. For each $i$ let $T_ i \subset |R_ i|$ be the set of points mapping to $(u_ i, u'_ i) \in |U_ i| \times |U_ i|$. This is a finite set by the definition of decent spaces (Decent Spaces, Definition 68.6.1). Moreover $T_ i$ is nonempty as we've assumed that $x$ and $x'$ map to the same point of $X_ i$. Hence $\mathop{\mathrm{lim}}\nolimits T_ i$ is nonempty, see Categories, Lemma 4.21.7. As before let $r \in |R| = \mathop{\mathrm{lim}}\nolimits |R_ i|$ be a point corresponding to an element of $\mathop{\mathrm{lim}}\nolimits T_ i$. Then $r$ maps to $(u, u')$ in $|U| \times |U|$ by construction and we see that $x = x'$ in $|X|$ as desired.

Parenthetical statement: A quasi-separated algebraic space is decent, see Decent Spaces, Section 68.6 (the key observation to this is Properties of Spaces, Lemma 66.6.7). A locally separated algebraic space is decent by Decent Spaces, Lemma 68.15.2. $\square$

Proof. We will use the criterion of Topology, Lemma 5.14.3. We have seen that $|X| = \mathop{\mathrm{lim}}\nolimits _ i |X_ i|$ as sets in Lemma 70.5.1. The maps $f_ i : X \to X_ i$ are morphisms of algebraic spaces hence determine continuous maps $|X| \to |X_ i|$. Thus $f_ i^{-1}(U_ i)$ is open for each open $U_ i \subset |X_ i|$. Finally, let $x \in |X|$ and let $x \in V \subset |X|$ be an open neighbourhood. We have to find an $i$ and an open neighbourhood $W_ i \subset |X_ i|$ of the image $x$ with $f_ i^{-1}(W_ i) \subset V$. Choose $0 \in I$. Choose a scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U = X \times _{X_0} U_0$ and $U_ i = X_ i \times _{X_0} U_0$ for $i \geq 0$. Then $U = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_ i$ in the category of schemes by Lemma 70.4.1. Choose $u \in U$ mapping to $x$. By the result for schemes (Limits, Lemma 32.4.2) we can find an $i \geq 0$ and an open neighbourhood $E_ i \subset U_ i$ of the image of $u$ whose inverse image in $U$ is contained in the inverse image of $V$ in $U$. Then we can set $W_ i \subset |X_ i|$ equal to the image of $E_ i$. This works because $|U_ i| \to |X_ i|$ is open. $\square$

Proof. Choose $0 \in I$. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = X_ i \times _{X_0} U_0$ and $U = X \times _{X_0} U_0$. Then each $U_ i$ is a nonempty affine scheme. Hence $U = \mathop{\mathrm{lim}}\nolimits U_ i$ is nonempty (Limits, Lemma 32.4.3) and thus $X$ is nonempty. $\square$

Proof. Set $Z = \overline{\{ x\} } \subset |X|$ and $Z_ i = \overline{\{ x_ i\} } \subset |X_ i|$. Since $|X| \to |X_ i|$ is continuous we see that $Z$ maps into $Z_ i$ for each $i$. Hence we obtain an injective map $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ because $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as sets (Lemma 70.5.1). Suppose that $x' \in |X|$ is not in $Z$. Then there is an open subset $U \subset |X|$ with $x' \in U$ and $x \not\in U$. Since $|X| = \mathop{\mathrm{lim}}\nolimits |X_ i|$ as topological spaces (Lemma 70.5.2) we can write $U = \bigcup _{j \in J} f_ j^{-1}(U_ j)$ for some subset $J \subset I$ and opens $U_ j \subset |X_ j|$, see Topology, Lemma 5.14.2. Then we see that for some $j \in J$ we have $f_ j(x') \in U_ j$ and $f_ j(x) \not\in U_ j$. In other words, we see that $f_ j(x') \not\in Z_ j$. Thus $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ as sets.

Next, endow $Z$ and $Z_ i$ with their reduced induced scheme structures, see Properties of Spaces, Definition 66.12.5. The transition morphisms $X_{i'} \to X_ i$ induce affine morphisms $Z_{i'} \to Z_ i$ and the projections $X \to X_ i$ induce compatible morphisms $Z \to Z_ i$. Hence we obtain morphisms $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i \to X$ of algebraic spaces. By Lemma 70.4.3 we see that $\mathop{\mathrm{lim}}\nolimits Z_ i \to X$ is a closed immersion. By Lemma 70.4.4 the algebraic space $\mathop{\mathrm{lim}}\nolimits Z_ i$ is reduced. By the above $Z \to \mathop{\mathrm{lim}}\nolimits Z_ i$ is bijective on points. By uniqueness of the reduced induced closed subscheme structure we find that this morphism is an isomorphism of algebraic spaces. $\square$

Proof. Choose a surjective étale morphism $U_0 \to X_0$ where $U_0$ is an affine scheme (Properties of Spaces, Lemma 66.6.3). Set $U_ i = X_ i \times _{X_0} U_0$. Set $R_0 = U_0 \times _{X_0} U_0$ and $R_ i = R_0 \times _{X_0} X_ i$. In the proof of Lemma 70.4.1 we have seen that there exists a presentation $X = U/R$ with $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and $R = \mathop{\mathrm{lim}}\nolimits R_ i$. Note that $U_ i$ and $U$ are affine and that $R_ i$ and $R$ are quasi-compact and separated (as $X_ i$ is quasi-separated). Hence Limits, Lemma 32.4.7 implies that

The lemma follows as $\Gamma (X, \mathcal{F}) = \mathop{\mathrm{Ker}}(\mathcal{F}(U) \to \mathcal{F}(R))$ and similarly $\Gamma (X_ i, \mathcal{F}_ i) = \mathop{\mathrm{Ker}}(\mathcal{F}_ i(U_ i) \to \mathcal{F}_ i(R_ i))$ $\square$

Proof. Follows formally from the construction of limits in Lemma 70.4.1 and the corresponding result for schemes: Limits, Lemma 32.4.11. $\square$

Proof. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = U_0 \times _{X_0} X_ i$ and $U = U_0 \times _{X_0} X$. Apply Limits, Lemma 32.8.11 to see that $Y_0 \times _{X_0} U_ i \to Z_0 \times _{X_0} U_ i$ is an isomorphism of schemes for some $i \geq 0$ (details omitted). As $U_ i \to X_ i$ is surjective étale, it follows that $Y_0 \times _{X_0} X_ i \to Z_0 \times _{X_0} X_ i$ is an isomorphism (details omitted). $\square$

Proof. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. For $i \geq 0$ set $U_ i = U_0 \times _{X_0} X_ i$ and set $U = U_0 \times _{X_0} X$. Note that $U_ i$ and $U$ are affine schemes which come equipped with surjective étale morphisms $U_ i \to X_ i$ and $U \to X$. Set $R_ i = U_ i \times _{X_ i} U_ i$ and $R = U \times _ X U$ with projections $s_ i, t_ i : R_ i \to U_ i$ and $s, t : R \to U$. Note that $R_ i$ and $R$ are quasi-compact separated schemes (as the algebraic spaces $X_ i$ and $X$ are quasi-separated). The maps $s_ i : R_ i \to U_ i$ and $s : R \to U$ are of finite type. By definition $X_ i$ is separated if and only if $(t_ i, s_ i) : R_ i \to U_ i \times U_ i$ is a closed immersion, and since $X$ is separated by assumption, the morphism $(t, s) : R \to U \times U$ is a closed immersion. Since $R \to U$ is of finite type, there exists an $i$ such that the morphism $R \to U_ i \times U$ is a closed immersion (Limits, Lemma 32.4.16). Fix such an $i \in I$. Apply Limits, Lemma 32.8.5 to the system of morphisms $R_{i'} \to U_ i \times U_{i'}$ for $i' \geq i$ (this is permissible as indeed $R_{i'} = R_ i \times _{U_ i \times U_ i} U_ i \times U_{i'}$) to see that $R_{i'} \to U_ i \times U_{i'}$ is a closed immersion for $i'$ sufficiently large. This implies immediately that $R_{i'} \to U_{i'} \times U_{i'}$ is a closed immersion finishing the proof of the lemma. $\square$

Proof. Choose $0 \in I$. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U = U_0 \times _{X_0} X$ and $U_ i = U_0 \times _{X_0} X_ i$ for $i \geq 0$. Since the transition morphisms are affine, the algebraic spaces $U_ i$ and $U$ are affine. Thus $U \to X$ is an étale morphism of affine schemes. Hence we can write $X = \mathop{\mathrm{Spec}}(A)$, $U = \mathop{\mathrm{Spec}}(B)$ and

such that $\Delta = \det (\partial g_\lambda /\partial x_\mu )$ is invertible in $B$, see Algebra, Lemma 10.143.2. Set $A_ i = \mathcal{O}_{X_ i}(X_ i)$. We have $A = \mathop{\mathrm{colim}}\nolimits A_ i$ by Lemma 70.5.6. After increasing $0$ we may assume we have $g_{1, i}, \ldots , g_{n, i} \in A_ i[x_1, \ldots , x_ n]$ mapping to $g_1, \ldots , g_ n$. Set

for all $i \geq 0$. Increasing $0$ if necessary we may assume that $\Delta _ i = \det (\partial g_{\lambda , i}/\partial x_\mu )$ is invertible in $B_ i$ for all $i \geq 0$. Thus $A_ i \to B_ i$ is an étale ring map. After increasing $0$ we may assume also that $\mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(A_ i)$ is surjective, see Limits, Lemma 32.8.15. Increasing $0$ yet again we may choose elements $h_{1, i}, \ldots , h_{n, i} \in \mathcal{O}_{U_ i}(U_ i)$ which map to the classes of $x_1, \ldots , x_ n$ in $B = \mathcal{O}_ U(U)$ and such that $g_{\lambda , i}(h_{\nu , i}) = 0$ in $\mathcal{O}_{U_ i}(U_ i)$. Thus we obtain a commutative diagram

By construction $B_ i = B_0 \otimes _{A_0} A_ i$ and $B = B_0 \otimes _{A_0} A$. Consider the morphism

This is a morphism of quasi-compact and quasi-separated algebraic spaces representable, separated and étale over $X_0$. The base change of $f_0$ to $X$ is an isomorphism by our choices. Hence Lemma 70.5.8 guarantees that there exists an $i$ such that the base change of $f_0$ to $X_ i$ is an isomorphism, in other words the diagram (70.5.10.1) is cartesian. Thus Descent, Lemma 35.39.1 applied to the fppf covering $\{ \mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(A_ i)\} $ combined with Descent, Lemma 35.37.1 give that $X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ is representable by a scheme affine over $\mathop{\mathrm{Spec}}(A_ i)$ as desired. (Of course it then also follows that $X_ i = \mathop{\mathrm{Spec}}(A_ i)$ but we don't need this.) $\square$

Proof. Choose a finite affine open covering $X = \bigcup W_ j$. By Lemma 70.5.7 we can find an $i \in I$ and open subspaces $W_{j, i} \subset X_ i$ whose base change to $X$ is $W_ j \to X$. By Lemma 70.5.10 we may assume that each $W_{j, i}$ is an affine scheme. This means that $X_ i$ is a scheme (see for example Properties of Spaces, Section 66.13). $\square$

Proof. Proof of (1). Choose $0 \in I$. As $X_0$ is quasi-compact, we can choose an affine scheme $W$ and an étale morphism $W \to B$ such that the image of $|X_0| \to |B|$ is contained in $|W| \to |B|$. Choose an affine scheme $U_0$ and an étale morphism $U_0 \to X_0 \times _ B W$ such that $U_0 \to X_0$ is surjective. (This is possible by our choice of $W$ and the fact that $X_0$ is quasi-compact; details omitted.) Let $V \to Y$, resp. $U \to X$, resp. $U_ i \to X_ i$ be the base change of $U_0 \to X_0$ (for $i \geq 0$). It suffices to prove that $V \to U_ i$ is a closed immersion for $i$ sufficiently large. Thus we reduce to proving the result for $V \to U = \mathop{\mathrm{lim}}\nolimits U_ i$ over $W$. This follows from the case of schemes, which is Limits, Lemma 32.4.16.

Proof of (2). Choose $0 \in I$. Choose a quasi-compact open subspace $X'_0 \subset X_0$ such that $Y \to X_0$ factors through $X'_0$. After replacing $X_ i$ by the inverse image of $X'_0$ for $i \geq 0$ we may assume all $X_ i'$ are quasi-compact and quasi-separated. Let $U \subset X$ be a quasi-compact open such that $Y \to X$ factors through a closed immersion $Y \to U$ ($U$ exists as $Y$ is quasi-compact). By Lemma 70.5.7 we may assume that $U = \mathop{\mathrm{lim}}\nolimits U_ i$ with $U_ i \subset X_ i$ quasi-compact open. By part (1) we see that $Y \to U_ i$ is a closed immersion for some $i$. Thus (2) holds.

Proof of (3). Choose $0 \in I$. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = X_ i \times _{X_0} U_0$, $U = X \times _{X_0} U_0 = Y \times _{X_0} U_0$. Then $U = \mathop{\mathrm{lim}}\nolimits U_ i$ is a limit of affine schemes, the transition maps of the system are closed immersions, and $U \to U_0$ is of finite presentation (because $U \to B$ is locally of finite presentation and $U_0 \to B$ is locally of finite type and Morphisms of Spaces, Lemma 67.28.9). Thus we've reduced to the following algebra fact: If $A = \mathop{\mathrm{lim}}\nolimits A_ i$ is a directed colimit of $R$-algebras with surjective transition maps and $A$ of finite presentation over $A_0$, then $A = A_ i$ for some $i$. Namely, write $A = A_0/(f_1, \ldots , f_ n)$. Pick $i$ such that $f_1, \ldots , f_ n$ map to zero under the surjective map $A_0 \to A_ i$.

Proof of (4). Set $Z_ i = Y \times _{X_ i} Y$. As the transition morphisms $X_{i'} \to X_ i$ are affine hence separated, the transition morphisms $Z_{i'} \to Z_ i$ are closed immersions, see Morphisms of Spaces, Lemma 67.4.5. We have $\mathop{\mathrm{lim}}\nolimits Z_ i = Y \times _ X Y = Y$ as $Y \to X$ is a monomorphism. Choose $0 \in I$. Since $Y \to X_0$ is locally of finite type (Morphisms of Spaces, Lemma 67.23.6) the morphism $Y \to Z_0$ is locally of finite presentation (Morphisms of Spaces, Lemma 67.28.10). The morphisms $Z_ i \to Z_0$ are locally of finite type (they are closed immersions). Finally, $Z_ i = Y \times _{X_ i} Y$ is quasi-compact as $X_ i$ is quasi-separated and $Y$ is quasi-compact. Thus part (3) applies to $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} Z_ i$ over $Z_0$ and we conclude $Y = Z_ i$ for some $i$. This proves (4) and the lemma. $\square$

Proof. Let $0 \in I$. Choose an affine scheme $W$ and an étale morphism $W \to Y$ such that the image of $|W| \to |Y|$ contains the image of $|X_0| \to |Y|$. This is possible as $X_0$ is quasi-compact. It suffices to check that $W \times _ Y X_ i \to W$ is separated for some $i \geq 0$ because the diagonal of $W \times _ Y X_ i$ over $W$ is the base change of $X_ i \to X_ i \times _ Y X_ i$ by the surjective étale morphism $(X_ i \times _ Y X_ i) \times _ Y W \to X_ i \times _ Y X_ i$. Since $Y$ is quasi-separated the algebraic spaces $W \times _ Y X_ i$ are quasi-compact (as well as quasi-separated). Thus we may base change to $W$ and assume $Y$ is an affine scheme. When $Y$ is an affine scheme, we have to show that $X_ i$ is a separated algebraic space for $i$ large enough and we are given that $X$ is a separated algebraic space. Thus this case follows from Lemma 70.5.9. $\square$

Proof. Choose an affine scheme $W$ and a surjective étale morphism $W \to Y$. Then $X \times _ Y W$ is affine and it suffices to check that $X_ i \times _ Y W$ is affine for some $i$ (Morphisms of Spaces, Lemma 67.20.3). This follows from Lemma 70.5.10. $\square$

Proof. Choose an affine scheme $W$ and a surjective étale morphism $W \to Y$. Then $X \times _ Y W$ is finite over $W$ and it suffices to check that $X_ i \times _ Y W$ is finite over $W$ for some $i$ (Morphisms of Spaces, Lemma 67.45.3). By Lemma 70.5.11 this reduces us to the case of schemes. In the case of schemes it follows from Limits, Lemma 32.4.19. $\square$

Proof. Choose an affine scheme $W$ and a surjective étale morphism $W \to Y$. Then $X \times _ Y W$ is a closed subspace of $W$ and it suffices to check that $X_ i \times _ Y W$ is a closed subspace $W$ for some $i$ (Morphisms of Spaces, Lemma 67.12.1). By Lemma 70.5.11 this reduces us to the case of schemes. In the case of schemes it follows from Limits, Lemma 32.4.20. $\square$