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54.4. Points of algebraic spaces

As is clear from Spaces, Example 53.14.8 a point of an algebraic space should not be defined as a monomorphism from the spectrum of a field. Instead we define them as equivalence classes of morphisms of spectra of fields exactly as explained in Schemes, Section 25.13.

Let $S$ be a scheme. Let $F$ be a presheaf on $(\textit{Sch}/S)_{fppf}$. Let $K$ is a field. Consider a morphism $$ \mathop{\rm Spec}(K) \longrightarrow F. $$ By the Yoneda Lemma this is given by an element $p \in F(\mathop{\rm Spec}(K))$. We say that two such pairs $(\mathop{\rm Spec}(K), p)$ and $(\mathop{\rm Spec}(L), q)$ are equivalent if there exists a third field $\Omega$ and a commutative diagram $$ \xymatrix{ \mathop{\rm Spec}(\Omega) \ar[r] \ar[d] & \mathop{\rm Spec}(L) \ar[d]^q \\ \mathop{\rm Spec}(K) \ar[r]^p & F. } $$ In other words, there are field extensions $K \to \Omega$ and $L \to \Omega$ such that $p$ and $q$ map to the same element of $F(\mathop{\rm Spec}(\Omega))$. We omit the verification that this defines an equivalence relation.

Definition 54.4.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. A point of $X$ is an equivalence class of morphisms from spectra of fields into $X$. The set of points of $X$ is denoted $|X|$.

Note that if $f : X \to Y$ is a morphism of algebraic spaces over $S$, then there is an induced map $|f| : |X| \to |Y|$ which maps a representative $x : \mathop{\rm Spec}(K) \to X$ to the representative $f \circ x : \mathop{\rm Spec}(K) \to Y$.

Lemma 54.4.2. Let $S$ be a scheme. Let $X$ be a scheme over $S$. The points of $X$ as a scheme are in canonical 1-1 correspondence with the points of $X$ as an algebraic space.

Proof. This is Schemes, Lemma 25.13.3. $\square$

Lemma 54.4.3. Let $S$ be a scheme. Let $$ \xymatrix{ Z \times_Y X \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y } $$ be a cartesian diagram of algebraic spaces over $S$. Then the map of sets of points $$ |Z \times_Y X| \longrightarrow |Z| \times_{|Y|} |X| $$ is surjective.

Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\rm Spec}(K) \to X$, $\mathop{\rm Spec}(L) \to Z$, then the assumption that they agree as elements of $|Y|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\mathop{\rm Spec}(M) \to \mathop{\rm Spec}(K) \to X \to Y$ and $\mathop{\rm Spec}(M) \to \mathop{\rm Spec}(L) \to Z \to Y$ agree. And this is exactly the condition which says you get a morphism $\mathop{\rm Spec}(M) \to Z \times_Y X$. $\square$

Lemma 54.4.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $f : T \to X$ be a morphism from a scheme to $X$. The following are equivalent

  1. $f : T \to X$ is surjective (according to Spaces, Definition 53.5.1), and
  2. $|f| : |T| \to |X|$ is surjective.

Proof. Assume (1). Let $x : \mathop{\rm Spec}(K) \to X$ be a morphism from the spectrum of a field into $X$. By assumption the morphism of schemes $\mathop{\rm Spec}(K) \times_X T \to \mathop{\rm Spec}(K)$ is surjective. Hence there exists a field extension $K \subset K'$ and a morphism $\mathop{\rm Spec}(K') \to \mathop{\rm Spec}(K) \times_X T$ such that the left square in the diagram $$ \xymatrix{ \mathop{\rm Spec}(K') \ar[r] \ar[d] & \mathop{\rm Spec}(K) \times_X T \ar[d] \ar[r] & T \ar[d] \\ \mathop{\rm Spec}(K) \ar@{=}[r] & \mathop{\rm Spec}(K) \ar[r]^-x & X } $$ is commutative. This shows that $|f| : |T| \to |X|$ is surjective.

Assume (2). Let $Z \to X$ be a morphism where $Z$ is a scheme. We have to show that the morphism of schemes $Z \times_X T \to T$ is surjective, i.e., that $|Z \times_X T| \to |Z|$ is surjective. This follows from (2) and Lemma 54.4.3. $\square$

Lemma 54.4.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $X = U/R$ be a presentation of $X$, see Spaces, Definition 53.9.3. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|X|$ is the quotient of $|U|$ by this equivalence relation.

Proof. The assumption means that $U$ is a scheme, $p : U \to X$ is a surjective, étale morphism, $R = U \times_X U$ is a scheme and defines an étale equivalence relation on $U$ such that $X = U/R$ as sheaves. By Lemma 54.4.4 we see that $|U| \to |X|$ is surjective. By Lemma 54.4.3 the map $$ |R| \longrightarrow |U| \times_{|X|} |U| $$ is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|X|$. Combining these two statements we get the result of the lemma. $\square$

Lemma 54.4.6. Let $S$ be a scheme. There exists a unique topology on the sets of points of algebraic spaces over $S$ with the following properties:

  1. for every morphism of algebraic spaces $X \to Y$ over $S$ the map $|X| \to |Y|$ is continuous, and
  2. for every étale morphism $U \to X$ with $U$ a scheme the map of topological spaces $|U| \to |X|$ is continuous and open.

Proof. Let $X$ be an algebraic space over $S$. Let $p : U \to X$ be a surjective étale morphism where $U$ is a scheme over $S$. We define $W \subset |X|$ is open if and only if $|p|^{-1}(W)$ is an open subset of $|U|$. This is a topology on $|X|$ (it is the quotient topology on $|X|$, see Topology, Lemma 5.6.2).

Let us prove that the topology is independent of the choice of the presentation. To do this it suffices to show that if $U'$ is a scheme, and $U' \to X$ is an étale morphism, then the map $|U'| \to |X|$ (with topology on $|X|$ defined using $U \to X$ as above) is open and continuous; which in addition will prove that (2) holds. Set $U'' = U \times_X U'$, so that we have the commutative diagram $$ \xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & X } $$ As $U \to X$ and $U' \to X$ are étale we see that both $U'' \to U$ and $U'' \to U'$ are étale morphisms of schemes. Moreover, $U'' \to U'$ is surjective. Hence we get a commutative diagram of maps of sets $$ \xymatrix{ |U''| \ar[r] \ar[d] & |U'| \ar[d] \\ |U| \ar[r] & |X| } $$ The lower horizontal arrow is surjective (see Lemma 54.4.4 or Lemma 54.4.5) and continuous by definition of the topology on $|X|$. The top horizontal arrow is surjective, continuous, and open by Morphisms, Lemma 28.34.13. The left vertical arrow is continuous and open (by Morphisms, Lemma 28.34.13 again.) Hence it follows formally that the right vertical arrow is continuous and open.

To finish the proof we prove (1). Let $a : X \to Y$ be a morphism of algebraic spaces. According to Spaces, Lemma 53.11.6 we can find a diagram $$ \xymatrix{ U \ar[d]_p \ar[r]_\alpha & V \ar[d]^q \\ X \ar[r]^a & Y } $$ where $U$ and $V$ are schemes, and $p$ and $q$ are surjective and étale. This gives rise to the diagram $$ \xymatrix{ |U| \ar[d]_p \ar[r]_\alpha & |V| \ar[d]^q \\ |X| \ar[r]^a & |Y| } $$ where all but the lower horizontal arrows are known to be continuous and the two vertical arrows are surjective and open. It follows that the lower horizontal arrow is continuous as desired. $\square$

Definition 54.4.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The underlying topological space of $X$ is the set of points $|X|$ endowed with the topology constructed in Lemma 54.4.6.

It turns out that this topological space carries the same information as the small Zariski site $X_{Zar}$ of Spaces, Definition 53.12.6.

Lemma 54.4.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.

  1. The rule $X' \mapsto |X'|$ defines an inclusion preserving bijection between open subspaces $X'$ (see Spaces, Definition 53.12.1) of $X$, and opens of the topological space $|X|$.
  2. A family $\{X_i \subset X\}_{i \in I}$ of open subspaces of $X$ is a Zariski covering (see Spaces, Definition 53.12.5) if and only if $|X| = \bigcup |X_i|$.

In other words, the small Zariski site $X_{Zar}$ of $X$ is canonically identified with a site associated to the topological space $|X|$ (see Sites, Example 7.6.4).

Proof. In order to prove (1) let us construct the inverse of the rule. Namely, suppose that $W \subset |X|$ is open. Choose a presentation $X = U/R$ corresponding to the surjective étale map $p : U \to X$ and étale maps $s, t : R \to U$. By construction we see that $|p|^{-1}(W)$ is an open of $U$. Denote $W' \subset U$ the corresponding open subscheme. It is clear that $R' = s^{-1}(W') = t^{-1}(W')$ is a Zariski open of $R$ which defines an étale equivalence relation on $W'$. By Spaces, Lemma 53.10.2 the morphism $X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space by Spaces, Lemma 53.11.3. By construction $|X'| = W$, i.e., $X'$ is a subspace of $X$ corresponding to $W$. Thus (1) is proved.

To prove (2), note that if $\{X_i \subset X\}_{i \in I}$ is a collection of open subspaces, then it is a Zariski covering if and only if the $U = \bigcup U \times_X X_i$ is an open covering. This follows from the definition of a Zariski covering and the fact that the morphism $U \to X$ is surjective as a map of presheaves on $(\textit{Sch}/S)_{fppf}$. On the other hand, we see that $|X| = \bigcup |X_i|$ if and only if $U = \bigcup U \times_X X_i$ by Lemma 54.4.5 (and the fact that the projections $U \times_X X_i \to X_i$ are surjective and étale). Thus the equivalence of (2) follows. $\square$

Lemma 54.4.9. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X' \subset X$ be an open subspace. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Then $f$ factors through $X'$ if and only if $|f| : |Y| \to |X|$ factors through $|X'| \subset |X|$.

Proof. By Spaces, Lemma 53.12.3 we see that $Y' = Y \times_X X' \to Y$ is an open immersion. If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by Lemma 54.4.8. $\square$

Lemma 54.4.10. Let $S$ be a scheme. Let $X$ be an algebraic spaces over $S$. Let $U$ be a scheme and let $f : U \to X$ be an étale morphism. Let $X' \subset X$ be the open subspace corresponding to the open $|f|(|U|) \subset |X|$ via Lemma 54.4.8. Then $f$ factors through a surjective étale morphism $f' : U \to X'$. Moreover, if $R = U \times_X U$, then $R = U \times_{X'} U$ and $X'$ has the presentation $X' = U/R$.

Proof. The existence of the factorization follows from Lemma 54.4.9. The morphism $f'$ is surjective according to Lemma 54.4.4. To see $f'$ is étale, suppose that $T \to X'$ is a morphism where $T$ is a scheme. Then $T \times_X U = T \times_{X'} U$ as $X" \to X$ is a monomorphism of sheaves. Thus the projection $T \times_{X'} U \to T$ is étale as we assumed $f$ étale. We have $U \times_X U = U \times_{X'} U$ as $X' \to X$ is a monomorphism. Then $X' = U/R$ follows from Spaces, Lemma 53.9.1. $\square$

Lemma 54.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map $$ \{\mathop{\rm Spec}(k) \to X \text{ monomorphism}\} \longrightarrow |X| $$ This map is injective.

Proof. Suppose that $\varphi_i : \mathop{\rm Spec}(k_i) \to X$ are monomorphisms for $i = 1, 2$. If $\varphi_1$ and $\varphi_2$ define the same point of $|X|$, then we see that the scheme $$ Y = \mathop{\rm Spec}(k_1) \times_{\varphi_1, X, \varphi_2} \mathop{\rm Spec}(k_2) $$ is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphisms $Y \to \mathop{\rm Spec}(k_i)$ are monomorphisms. Hence $\mathop{\rm Spec}(k_1) = Y = \mathop{\rm Spec}(k_2)$ as schemes over $X$, see Schemes, Lemma 25.23.10. We conclude that $\varphi_1 = \varphi_2$, which proves the lemma. $\square$

We will see in Decent Spaces, Lemma 56.11.1 that this map is a bijection when $X$ is decent.

    The code snippet corresponding to this tag is a part of the file spaces-properties.tex and is located in lines 217–607 (see updates for more information).

    \section{Points of algebraic spaces}
    \label{section-points}
    
    \noindent
    As is clear from
    Spaces, Example \ref{spaces-example-affine-line-translation}
    a point of an algebraic space should not be defined as a monomorphism
    from the spectrum of a field.
    Instead we define them as equivalence classes of morphisms of spectra
    of fields exactly as explained in
    Schemes, Section \ref{schemes-section-points}.
    
    \medskip\noindent
    Let $S$ be a scheme.
    Let $F$ be a presheaf on $(\Sch/S)_{fppf}$.
    Let $K$ is a field. Consider a morphism
    $$
    \Spec(K) \longrightarrow F.
    $$
    By the Yoneda Lemma this is given by an
    element $p \in F(\Spec(K))$. We say that two such
    pairs $(\Spec(K), p)$ and $(\Spec(L), q)$
    are {\it equivalent} if there exists
    a third field $\Omega$ and a commutative diagram
    $$
    \xymatrix{
    \Spec(\Omega) \ar[r] \ar[d] &
    \Spec(L) \ar[d]^q \\
    \Spec(K) \ar[r]^p &
    F.
    }
    $$
    In other words, there are field extensions
    $K \to \Omega$ and $L \to \Omega$ such that
    $p$ and $q$ map to the same element
    of $F(\Spec(\Omega))$. We omit the verification that this
    defines an equivalence relation.
    
    \begin{definition}
    \label{definition-points}
    Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
    A {\it point} of $X$ is an equivalence class of morphisms
    from spectra of fields into $X$.
    The set of points of $X$ is denoted $|X|$.
    \end{definition}
    
    \noindent
    Note that if $f : X \to Y$ is a morphism of algebraic spaces
    over $S$, then there is an induced map $|f| : |X| \to |Y|$ which
    maps a representative $x : \Spec(K) \to X$ to the representative
    $f \circ x : \Spec(K) \to Y$.
    
    \begin{lemma}
    \label{lemma-scheme-points}
    Let $S$ be a scheme. Let $X$ be a scheme over $S$.
    The points of $X$ as a scheme are in canonical 1-1 correspondence
    with the points of $X$ as an algebraic space.
    \end{lemma}
    
    \begin{proof}
    This is Schemes, Lemma \ref{schemes-lemma-characterize-points}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-points-cartesian}
    Let $S$ be a scheme. Let
    $$
    \xymatrix{
    Z \times_Y X \ar[r] \ar[d] & X \ar[d] \\
    Z \ar[r] & Y
    }
    $$
    be a cartesian diagram of algebraic spaces over $S$. Then the map of sets
    of points
    $$
    |Z \times_Y X|
    \longrightarrow
    |Z| \times_{|Y|} |X|
    $$
    is surjective.
    \end{lemma}
    
    \begin{proof}
    Namely, suppose given fields $K$, $L$ and morphisms
    $\Spec(K) \to X$, $\Spec(L) \to Z$, then the
    assumption that they agree as elements of $|Y|$ means that
    there is a common extension $K \subset M$ and $L \subset M$
    such that
    $\Spec(M) \to \Spec(K) \to X \to Y$ and
    $\Spec(M) \to \Spec(L) \to Z \to Y$ agree.
    And this is exactly the condition which says you get a
    morphism $\Spec(M) \to Z \times_Y X$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-characterize-surjective}
    Let $S$ be a scheme.
    Let $X$ be an algebraic space over $S$.
    Let $f : T \to X$ be a morphism from a scheme to $X$.
    The following are equivalent
    \begin{enumerate}
    \item $f : T \to X$ is surjective (according to
    Spaces, Definition \ref{spaces-definition-relative-representable-property}),
    and
    \item $|f| : |T| \to |X|$ is surjective.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Assume (1). Let $x : \Spec(K) \to X$ be a morphism
    from the spectrum of a field into $X$. By assumption the morphism of
    schemes $\Spec(K) \times_X T \to \Spec(K)$ is surjective.
    Hence there exists a field extension $K \subset K'$ and a morphism
    $\Spec(K') \to \Spec(K) \times_X T$ such that the left
    square in the diagram
    $$
    \xymatrix{
    \Spec(K') \ar[r] \ar[d] &
    \Spec(K) \times_X T \ar[d] \ar[r] &
    T \ar[d] \\
    \Spec(K) \ar@{=}[r] &
    \Spec(K) \ar[r]^-x & X
    }
    $$
    is commutative. This shows that $|f| : |T| \to |X|$ is surjective.
    
    \medskip\noindent
    Assume (2). Let $Z \to X$ be a morphism where $Z$ is
    a scheme. We have to show that the morphism of schemes $Z \times_X T \to T$
    is surjective, i.e., that $|Z \times_X T| \to |Z|$ is surjective.
    This follows from (2) and
    Lemma \ref{lemma-points-cartesian}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-points-presentation}
    Let $S$ be a scheme.
    Let $X$ be an algebraic space over $S$.
    Let $X = U/R$ be a presentation of $X$, see
    Spaces, Definition \ref{spaces-definition-presentation}.
    Then the image of $|R| \to |U| \times |U|$ is an equivalence relation
    and $|X|$ is the quotient of $|U|$ by this equivalence relation.
    \end{lemma}
    
    \begin{proof}
    The assumption means that $U$ is a scheme, $p : U \to X$ is a surjective,
    \'etale morphism, $R = U \times_X U$ is a scheme and defines an \'etale
    equivalence relation on $U$ such that $X = U/R$ as sheaves. By
    Lemma \ref{lemma-characterize-surjective}
    we see that $|U| \to |X|$ is surjective. By
    Lemma \ref{lemma-points-cartesian}
    the map
    $$
    |R| \longrightarrow |U| \times_{|X|} |U|
    $$
    is surjective. Hence the image of $|R| \to |U| \times |U|$ is
    exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$
    such that $u_1$ and $u_2$ have the same image in $|X|$.
    Combining these two statements we get the result of the lemma.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-topology-points}
    Let $S$ be a scheme. There exists a unique topology on the sets of points
    of algebraic spaces over $S$ with the following properties:
    \begin{enumerate}
    \item for every morphism of algebraic spaces $X \to Y$ over $S$
    the map $|X| \to |Y|$ is continuous, and
    \item for every \'etale morphism $U \to X$ with $U$ a scheme
    the map of topological spaces $|U| \to |X|$ is continuous and open.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Let $X$ be an algebraic space over $S$. Let $p : U \to X$ be a
    surjective \'etale morphism where $U$ is a scheme over $S$.
    We define $W \subset |X|$ is open if and only if $|p|^{-1}(W)$
    is an open subset of $|U|$. This is a topology on $|X|$
    (it is the quotient topology on $|X|$, see
    Topology, Lemma \ref{topology-lemma-quotient}).
    
    \medskip\noindent
    Let us prove that the topology is independent of the choice of
    the presentation. To do this it suffices to show that if $U'$ is a scheme,
    and $U' \to X$ is an \'etale morphism, then the map $|U'| \to |X|$
    (with topology on $|X|$ defined using $U \to X$ as above)
    is open and continuous; which in addition will prove that (2) holds.
    Set $U'' = U \times_X U'$, so that we have the commutative diagram
    $$
    \xymatrix{
    U'' \ar[r] \ar[d] & U' \ar[d] \\
    U \ar[r] & X
    }
    $$
    As $U \to X$ and $U' \to X$ are \'etale we see that
    both $U'' \to U$ and $U'' \to U'$ are \'etale morphisms of schemes.
    Moreover, $U'' \to U'$ is surjective. Hence
    we get a commutative diagram of maps of sets
    $$
    \xymatrix{
    |U''| \ar[r] \ar[d] & |U'| \ar[d] \\
    |U| \ar[r] & |X|
    }
    $$
    The lower horizontal arrow is surjective (see
    Lemma \ref{lemma-characterize-surjective}
    or
    Lemma \ref{lemma-points-presentation})
    and continuous by definition of the topology on $|X|$.
    The top horizontal arrow is surjective, continuous, and open by
    Morphisms, Lemma \ref{morphisms-lemma-etale-open}.
    The left vertical arrow is continuous and open (by
    Morphisms, Lemma \ref{morphisms-lemma-etale-open}
    again.) Hence it follows formally that the right vertical
    arrow is continuous and open.
    
    \medskip\noindent
    To finish the proof we prove (1).
    Let $a : X \to Y$ be a morphism of algebraic spaces. According to
    Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}
    we can find a diagram
    $$
    \xymatrix{
    U \ar[d]_p \ar[r]_\alpha & V \ar[d]^q \\
    X \ar[r]^a & Y
    }
    $$
    where $U$ and $V$ are schemes, and $p$ and $q$ are surjective and \'etale.
    This gives rise to the diagram
    $$
    \xymatrix{
    |U| \ar[d]_p \ar[r]_\alpha & |V| \ar[d]^q \\
    |X| \ar[r]^a & |Y|
    }
    $$
    where all but the lower horizontal arrows are known to be continuous and
    the two vertical arrows are surjective and open. It follows that the
    lower horizontal arrow is continuous as desired.
    \end{proof}
    
    \begin{definition}
    \label{definition-topological-space}
    Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
    The underlying {\it topological space} of $X$ is the set of points
    $|X|$ endowed with the topology constructed in
    Lemma \ref{lemma-topology-points}.
    \end{definition}
    
    \noindent
    It turns out that this topological space carries the same information
    as the small Zariski site $X_{Zar}$ of
    Spaces, Definition \ref{spaces-definition-small-Zariski-site}.
    
    \begin{lemma}
    \label{lemma-open-subspaces}
    Let $S$ be a scheme.
    Let $X$ be an algebraic space over $S$.
    \begin{enumerate}
    \item The rule $X' \mapsto |X'|$ defines an inclusion preserving
    bijection between open subspaces $X'$ (see
    Spaces, Definition \ref{spaces-definition-immersion})
    of $X$, and opens of the topological space $|X|$.
    \item A family $\{X_i \subset X\}_{i \in I}$ of open subspaces of $X$
    is a Zariski covering (see
    Spaces, Definition \ref{spaces-definition-Zariski-open-covering})
    if and only if $|X| = \bigcup |X_i|$.
    \end{enumerate}
    In other words, the small Zariski site $X_{Zar}$ of $X$ is canonically
    identified with a site associated to the topological space $|X|$ (see
    Sites, Example \ref{sites-example-site-topological}).
    \end{lemma}
    
    \begin{proof}
    In order to prove (1) let us construct the inverse of the rule.
    Namely, suppose that $W \subset |X|$ is open. Choose a presentation
    $X = U/R$ corresponding to the surjective \'etale map
    $p : U \to X$ and \'etale maps $s, t : R \to U$.
    By construction we see that $|p|^{-1}(W)$ is an
    open of $U$. Denote $W' \subset U$ the corresponding open subscheme.
    It is clear that $R' = s^{-1}(W') = t^{-1}(W')$ is a Zariski open
    of $R$ which defines an \'etale equivalence relation on $W'$.
    By Spaces, Lemma \ref{spaces-lemma-finding-opens} the morphism
    $X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space
    by Spaces, Lemma \ref{spaces-lemma-representable-over-space}.
    By construction $|X'| = W$, i.e., $X'$ is a subspace of $X$
    corresponding to $W$. Thus (1) is proved.
    
    \medskip\noindent
    To prove (2), note that if $\{X_i \subset X\}_{i \in I}$ is a collection
    of open subspaces, then it is a Zariski covering if and only if the
    $U = \bigcup U \times_X X_i$ is an open covering. This follows from
    the definition of a Zariski covering and the fact that the morphism
    $U \to X$ is surjective as a map of presheaves on $(\Sch/S)_{fppf}$.
    On the other hand, we see that $|X| = \bigcup |X_i|$ if and only if
    $U = \bigcup U \times_X X_i$ by Lemma \ref{lemma-points-presentation}
    (and the fact that the projections $U \times_X X_i \to X_i$ are surjective
    and \'etale). Thus the equivalence of (2) follows.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-factor-through-open-subspace}
    Let $S$ be a scheme.
    Let $X$, $Y$ be algebraic spaces over $S$.
    Let $X' \subset X$ be an open subspace.
    Let $f : Y \to X$ be a morphism of algebraic spaces over $S$.
    Then $f$ factors through $X'$ if and only if $|f| : |Y| \to |X|$
    factors through $|X'| \subset |X|$.
    \end{lemma}
    
    \begin{proof}
    By Spaces, Lemma \ref{spaces-lemma-base-change-immersions}
    we see that $Y' = Y \times_X X' \to Y$ is an open immersion.
    If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by
    Lemma \ref{lemma-open-subspaces}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-etale-image-open}
    Let $S$ be a scheme. Let $X$ be an algebraic spaces over $S$.
    Let $U$ be a scheme and let $f : U \to X$ be an \'etale morphism.
    Let $X' \subset X$ be the open subspace corresponding to
    the open $|f|(|U|) \subset |X|$ via
    Lemma \ref{lemma-open-subspaces}.
    Then $f$ factors through a surjective \'etale morphism $f' : U \to X'$.
    Moreover, if $R = U \times_X U$, then $R = U \times_{X'} U$ and $X'$ has
    the presentation $X' = U/R$.
    \end{lemma}
    
    \begin{proof}
    The existence of the factorization follows from
    Lemma \ref{lemma-factor-through-open-subspace}.
    The morphism $f'$ is surjective according to
    Lemma \ref{lemma-characterize-surjective}.
    To see $f'$ is \'etale, suppose that $T \to X'$ is a morphism
    where $T$ is a scheme. Then $T \times_X U = T \times_{X'} U$
    as $X" \to X$ is a monomorphism of sheaves. Thus the projection
    $T \times_{X'} U \to T$ is \'etale as we assumed $f$ \'etale.
    We have $U \times_X U = U \times_{X'} U$ as $X' \to X$ is a monomorphism.
    Then $X' = U/R$ follows from
    Spaces, Lemma \ref{spaces-lemma-space-presentation}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-points-monomorphism}
    Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
    Consider the map
    $$
    \{\Spec(k) \to X \text{ monomorphism}\}
    \longrightarrow
    |X|
    $$
    This map is injective.
    \end{lemma}
    
    \begin{proof}
    Suppose that $\varphi_i : \Spec(k_i) \to X$ are monomorphisms
    for $i = 1, 2$. If $\varphi_1$ and $\varphi_2$ define the same point
    of $|X|$, then we see that the scheme
    $$
    Y = \Spec(k_1) \times_{\varphi_1, X, \varphi_2} \Spec(k_2)
    $$
    is nonempty. Since the base change of a monomorphism is a monomorphism
    this means that the projection morphisms $Y \to \Spec(k_i)$
    are monomorphisms. Hence $\Spec(k_1) = Y = \Spec(k_2)$
    as schemes over $X$, see
    Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}.
    We conclude that $\varphi_1 = \varphi_2$, which proves the lemma.
    \end{proof}
    
    \noindent
    We will see in
    Decent Spaces,
    Lemma \ref{decent-spaces-lemma-decent-points-monomorphism}
    that this map is a bijection when $X$ is decent.

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