From 2b97b81fdcaf2d849ca6fe708f97cd73590f83c5 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Thu, 8 Jun 2023 15:15:09 +0200
Subject: [PATCH 1/9] Simplified proofs
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

- Changes in categories.tex:
· Added lemma-pasting-law-pullbacks, https://stacks.math.columbia.edu/tag/001U#comment-3413
· Added lemma-3x3-pullback, from which the newly added lemma-magic-square and lemma-base-change-diagonal-1-cat easily follow (as well as one direction in the proof of lemma-pasting-law-pullbacks).

- Changes in schemes.tex
· Added changes proposed in https://stacks.math.columbia.edu/tag/01JW#comment-8464 and https://stacks.math.columbia.edu/tag/01JY#comment-8466
· Rewrote proof of lemma-fibre-product-after-map by referencing lemma-magic-square.
· Simplified proof of (3) and (4) in lemma-separated-permanence by referencing lemma-base-change-diagonal-1-cat.
---
 categories.tex | 121 +++++++++++++++++++++++++++++++++++++++++++++++++
 schemes.tex    | 107 ++++++++++++++++++-------------------------
 2 files changed, 166 insertions(+), 62 deletions(-)

diff --git a/categories.tex b/categories.tex
index 6cacb0be8..d9d22c4e3 100644
--- a/categories.tex
+++ b/categories.tex
@@ -755,6 +755,127 @@ \section{Fibre products}
 and $g\in \Mor_{\mathcal C}(z, y)$.
 \end{definition}
 
+\begin{proof}
+\label{definition-diagonal-morphism}
+Let $\mathcal{C}$ be a category, let $f:x\to y$ be a
+morphism in $\mathcal{C}$, and suppose that $x\times_y x$ exists.
+The {\it diagonal morphism}, denoted $\Delta_f$ or $\Delta_{x/y}$,
+is the unique morphism $x\to x\times_y x$ coming from the identity $x\to x$.
+\end{proof}
+
+\begin{lemma}
+\begin{slogan}
+The horizontal fibre product of the vertical fibre products
+equals the vertical fibre product of the horizontal fibre products.
+\end{slogan}
+\label{lemma-3x3-pullback}
+Let $\mathcal{C}$ be a category that has fibre products and suppose that
+$$
+\xymatrix{
+X_1 \ar[r]\ar[d] & X_0\ar[d] & X_2 \ar[l]\ar[d]\\
+S_1 \ar[r] & S_0 & S_2 \ar[l] \\
+Y_1 \ar[r]\ar[u] & Y_0\ar[u] & Y_2 \ar[l]\ar[u]
+}
+$$
+is a commutative diagram in $\mathcal{C}$. Then we have
+$$
+(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2)
+= (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2).
+$$
+\end{lemma}
+
+\begin{proof}
+First proof: Use the fact that the Yoneda embedding
+$\mathcal{C}\to\text{Fun}(\mathcal{C}^\text{opp},Sets)$
+preserves limits to reduce it to the case of $Sets$,
+where it is easier to check the two sets are isomorphic.
+
+\medskip\noindent
+Second proof: Define a pair of anti-parallel canonical maps
+between the two sides of the equality from the statement.
+Do so leveraging the universal properties of all
+the involved fibre products.
+Verify that these maps are mutually inverse.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-magic-square}
+Let $\mathcal{C}$ be a category that has fibre products.
+Let $X\to S$, $Y\to S$, $S\to T$ be morphisms in $\mathcal{C}$.
+There is a cartesian diagram
+$$
+\xymatrix{
+X \times_T Y \ar[r] \ar[d] & X \times_S Y \ar[d] \\
+T \ar[r]^(.4){\Delta_{T/S}} \ar[r] & T \times_S T
+}
+$$
+\end{lemma}
+
+\begin{proof}
+Apply Lemma \ref{lemma-3x3-pullback} to the diagram
+$$
+\xymatrix{
+S \ar[r]\ar[d] & S \ar[d] & X \ar[l]\ar[d]\\
+S \ar[r]       & T        & T \ar[l] \\
+S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
+}
+$$
+\end{proof}
+
+\begin{lemma}
+\label{lemma-base-change-diagonal-1-cat}
+Let $\mathcal{C}$ be a category with fibre products.
+Let $X\to S$, $S'\to S$ be morphisms in $\mathcal{C}$,
+and let $X'=S'\times_SX$. There is a cartesian square
+$$
+\xymatrix{
+X' \ar[r]^(.37){\Delta_{X'/S'}}\ar[d] & X'\times_{S'}X'\ar[d]\\
+X \ar[r]^(.35){\Delta_{X/S}} & X\times_{S}X
+}
+$$
+\end{lemma}
+
+\begin{proof}
+Apply Lemma \ref{lemma-3x3-pullback} to the diagram
+$$
+\xymatrix{
+X \ar[r]\ar[d] & X \ar[d] & X' \ar[l]\ar[d]\\
+X \ar[r]       & S        & S' \ar[l] \\
+X \ar[r]\ar[u] & X \ar[u] & X' \ar[l]\ar[u]
+}
+$$
+\end{proof}
+
+\begin{lemma}
+	\label{lemma-pasting-law-pullbacks}
+	Let
+	$$
+	\xymatrix{
+		X' \ar@{->}[r] \ar@{->}[d] & X \ar@{->}[d] \\
+		Y' \ar@{->}[r] \ar@{->}[d] & T \ar@{->}[d] \\
+		Y \ar@{->}[r] & S
+	}
+	$$
+	be a commutative diagram in a category,
+	and suppose that the lower square is cartesian.
+	Then the outer rectangle is cartesian
+	if and only if the upper square is cartesian.
+\end{lemma}
+
+\begin{proof}
+	Suppose each inner square is cartesian. Then the outer rectangle is cartesian by Lemma \ref{lemma-3x3-pullback} applied to the diagram
+	$$
+	\xymatrix{
+		X \ar[r]\ar[d] & T \ar[d] & T \ar[l]\ar[d]\\
+		S \ar[r]       & S        & S \ar[l] \\
+		S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
+	}
+	$$
+	
+	\medskip\noindent
+	We omit the proof of the other implication.
+\end{proof}
+
 \begin{definition}
 \label{definition-representable-morphism}
 A morphism $f : x \to y$ of a category $\mathcal{C}$ is said to be
diff --git a/schemes.tex b/schemes.tex
index 759943ccc..80e44d828 100644
--- a/schemes.tex
+++ b/schemes.tex
@@ -3438,36 +3438,6 @@ \section{Base change in algebraic geometry}
 \end{enumerate}
 \end{definition}
 
-\noindent
-Here is a typical result.
-
-\begin{lemma}
-\label{lemma-base-change-immersion}
-Let $S$ be a scheme. Let $f : X \to Y$ be an
-immersion (resp.\ closed immersion, resp. open immersion)
-of schemes over $S$. Then any base change of $f$ is an
-immersion (resp.\ closed immersion, resp. open immersion).
-\end{lemma}
-
-\begin{proof}
-We can think of the base change of $f$ via the morphism
-$S' \to S$ as the top left vertical arrow in the following
-commutative diagram:
-$$
-\xymatrix{
-X_{S'} \ar[r] \ar[d] & X \ar[d] \ar@/^4ex/[dd] \\
-Y_{S'} \ar[r] \ar[d] & Y \ar[d] \\
-S' \ar[r] & S
-}
-$$
-The diagram implies $X_{S'} \cong Y_{S'} \times_Y X$,
-and the lemma follows from Lemma \ref{lemma-fibre-product-immersion}.
-\end{proof}
-
-\noindent
-In fact this type of result is so typical that there is a
-piece of language to express it. Here it is.
-
 \begin{definition}
 \label{definition-preserved-by-base-change}
 Properties and base change.
@@ -3485,7 +3455,43 @@ \section{Base change in algebraic geometry}
 \end{enumerate}
 \end{definition}
 
-\noindent
+This terminology never gives rise to ambiguities.
+
+\begin{remark}
+\label{remark-preserved-by-base-change}
+Let $\mathcal{P}$ be a property of morphisms of schemes.
+Then $\mathcal{P}$ is preserved under arbitrary base change
+as in \ref{definition-preserved-by-base-change} (1) if and
+only if it is preserved under arbitrary base change as in
+\ref{definition-preserved-by-base-change} (2).
+To see ($\Rightarrow$), use the diagram
+$$
+\xymatrix{
+	X_{S'} \ar[r] \ar[d] & X \ar[d] \ar@/^4ex/[dd] \\
+	Y_{S'} \ar[r] \ar[d] & Y \ar[d] \\
+	S' \ar[r] & S
+}
+$$
+where the top square is cartesian by
+Categories, Lemma \ref{lemma-pasting-law-pullbacks}.
+To see ($\Leftarrow$), use again the same diagram
+but with $Y=S$, so $Y_{S'}=S'$.
+\end{remark}
+
+Here is a typical result.
+
+\begin{lemma}
+	\label{lemma-base-change-immersion}
+	Let $S$ be a scheme. Let $f : X \to Y$ be an
+	immersion (resp.\ closed immersion, resp. open immersion)
+	of schemes over $S$. Then any base change of $f$ is an
+	immersion (resp.\ closed immersion, resp. open immersion).
+\end{lemma}
+
+\begin{proof}
+	Use Remark \ref{remark-preserved-by-base-change} and Lemma \ref{lemma-fibre-product-immersion}.
+\end{proof}
+
 At this point we can say that ``being a closed immersion'' is
 preserved under arbitrary base change.
 
@@ -4221,15 +4227,9 @@ \section{Separation axioms}
 \end{lemma}
 
 \begin{proof}
-By general category theory the following diagram
-$$
-\xymatrix{
-X \times_T Y \ar[r] \ar[d] & X \times_S Y \ar[d] \\
-T \ar[r]^{\Delta_{T/S}} \ar[r] & T \times_S T
-}
-$$
-is a fibre product diagram. The lemma follows
-from Lemmas \ref{lemma-diagonal-immersion},
+Use the cartesianity of the square from Categories,
+Lemma \ref{lemma-magic-square} and apply
+Lemmas \ref{lemma-diagonal-immersion},
 \ref{lemma-fibre-product-immersion} and
 \ref{lemma-quasi-compact-preserved-base-change}.
 \end{proof}
@@ -4285,28 +4285,11 @@ \section{Separation axioms}
 works for ``quasi-separated'' (with the same references).
 
 \medskip\noindent
-Let $f : X \to Y$ be a morphism of schemes over a base $S$.
-Let $S' \to S$ be a morphism of schemes. Let $f' : X_{S'} \to Y_{S'}$
-be the base change of $f$. Then the diagonal morphism
-of $f'$ is a morphism
-$$
-\Delta_{f'} :
-X_{S'} = S' \times_S X
-\longrightarrow
-X_{S'} \times_{Y_{S'}} X_{S'} = S' \times _S (X \times_Y X)
-$$
-which is easily seen to be the base change of $\Delta_f$.
-Thus (3) and (4) follow from the fact that
-closed immersions and quasi-compact morphisms are preserved
-under arbitrary base change (Lemmas
-\ref{lemma-fibre-product-immersion} and
-\ref{lemma-quasi-compact-preserved-base-change}).
-
-\medskip\noindent
-If $f : X \to Y$ and $g : U \to V$ are morphisms of schemes over a base $S$,
-then $f \times g$ is the composition of $X \times_S U \to X \times_S V$
-(a base change of $g$) and $X \times_S V \to Y \times_S V$ (a base change
-of $f$). Hence (5) and (6) follow from (1) -- (4).
+(3) and (4) follow from Categories,
+Lemma \ref{lemma-base-change-diagonal-1-cat} plus
+Lemmas \ref{lemma-fibre-product-immersion} and
+\ref{lemma-quasi-compact-preserved-base-change}.
+(5) and (6) are a particular case of (3) and (4).
 \end{proof}
 
 \begin{lemma}

From b37dfcf7631b118dd4534336d5862518c40b3b30 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Mon, 12 Jun 2023 16:22:15 +0200
Subject: [PATCH 2/9] characterizations of (quasi-)separatedness
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

The aim of this commit is to incorporate the result "(quasi-)separatedness is local on the target" and propositions (1.2.6) and (1.2.7) of EGA IV, première partie (they are in pp. 227-228). (All of them into the section Separation axioms on the Schemes chapter.)

The first one is quasi-separated-local-target. The proof introduces a lot of single-use definitions regarding properties of morphisms of schemes, and actually shows a general result for these definitions (P is LOCT implies Δ_P is LOCT). Such a phrasing for a proof might look weird. I would have written a new section instead on the Schemes chapter with the new definitions plus the mentioned result, but there's been objections to this: https://stacks.math.columbia.edu/tag/01KH#comment-7560

I also changed slightly the statement of lemma-closed-local-target and the statement and proof of lemma-quasi-compact-affine to have the results "«being quasi-compact» is LOCT" and "«being a closed immersion» is LOCT" fully spelled.

Propositions (1.2.6) and (1.2.7) of EGA IV correspond to lemma-characterize-quasi-separated-scheme and lemma-quasi-separatedness-is-topological. In their statements, I've introduced the modern terminology of "quasi-separated topological space" and "quasi-separated map of topological spaces" (EGA doesn't use them), defined in a newly-added section in Topology. There's been hesitations to the introduction of such a section before (see comments in tag 0067). Nonetheless, I thought that giving the concept a name and using it makes more manifest the topological nature of quasi-separatedness. If you think this is not a sufficient reason to include the new section in Topology (I understand the reasons in your comment in tag 0067), one could instead fully spell out the concept each time it appears in lemma-characterize-quasi-separated-scheme and lemma-quasi-separatedness-is-topological.

(Note: the statement of Proposition (1.2.7) of EGA IV contains an erratum, see https://math.stackexchange.com/q/4716228/394668 )
---
 categories.tex |   3 +
 schemes.tex    | 179 +++++++++++++++++++++++++++++++++++++++++++------
 topology.tex   |  26 +++++++
 3 files changed, 189 insertions(+), 19 deletions(-)

diff --git a/categories.tex b/categories.tex
index d9d22c4e3..755e11973 100644
--- a/categories.tex
+++ b/categories.tex
@@ -824,6 +824,9 @@ \section{Fibre products}
 
 \begin{lemma}
 \label{lemma-base-change-diagonal-1-cat}
+\begin{slogan}
+	The base change of the diagonal is the diagonal of the base change.
+\end{slogan}
 Let $\mathcal{C}$ be a category with fibre products.
 Let $X\to S$, $S'\to S$ be morphisms in $\mathcal{C}$,
 and let $X'=S'\times_SX$. There is a cartesian square
diff --git a/schemes.tex b/schemes.tex
index 80e44d828..c37ac483b 100644
--- a/schemes.tex
+++ b/schemes.tex
@@ -356,7 +356,9 @@ \section{Closed immersions of locally ringed spaces}
 \begin{lemma}
 \label{lemma-closed-local-target}
 Let $f : Z \to X$ be a morphism of locally ringed spaces.
-In order for $f$ to be a closed immersion it suffices
+Then $f|_{f^{-1}U}:f^{-1}U\to U$ is a closed immersion
+for all open subsets $U\subset X$.
+Moreover, in order for $f$ to be a closed immersion it suffices
 that there exists an open covering $X = \bigcup U_i$ such
 that each $f : f^{-1}U_i \to U_i$ is a closed immersion.
 \end{lemma}
@@ -3571,19 +3573,44 @@ \section{Quasi-compact morphisms}
 The following are equivalent
 \begin{enumerate}
 \item $f : X \to S$ is quasi-compact,
-\item the inverse image of every affine open is quasi-compact, and
+\item the inverse image of every affine open is quasi-compact,
+\item there is an open cover $S=\bigcup_{i\in I} U_i$
+such that the restriction $f^{-1}(U_i)\to U_i$
+is quasi-compact for all $i$, and
 \item there exists some affine open covering $S = \bigcup_{i \in I} U_i$
 such that $f^{-1}(U_i)$ is quasi-compact for all $i$.
 \end{enumerate}
 \end{lemma}
 
 \begin{proof}
+(1)$\Rightarrow$(2).
+Affine schemes are quasi-compact,
+see Algebra, Lemma \ref{lemma-quasi-compact}.
+
+\medskip\noindent
+(2)$\Rightarrow$(1).
+Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis
+of the topology consisting of affine opens we see that $K$ is a finite
+union of affine opens. Hence the inverse image of $K$ is a finite
+union of affine opens. Hence $f$ is quasi-compact.
+
+\medskip\noindent
+(1)$\Rightarrow$(3). Trivial.
+
+\medskip\noindent
+(3)$\Rightarrow$(4).
+For each $i\in I$, take an open affine cover $U_i=\bigcup_{k\in K_i}V_{ik}$,
+so $f^{-1}(V_{ik})$ is quasi-compact.
+The cover $Y=\bigcup_{i\in I}\bigcup_{k\in K_i}V_{ik}$ suffices.
+
+\medskip\noindent
+(4)$\Rightarrow$(1).
 Suppose we are given a covering $S = \bigcup_{i \in I} U_i$ as in (3).
-First, let $U \subset S$ be any affine open. For any $u \in U$
+First, let $U \subset S$ be any quasi-compact open subset. For any $u \in U$
 we can find an index $i(u) \in I$ such that $u \in U_{i(u)}$.
 As standard opens form a basis for the topology on $U_{i(u)}$ we can find
-$W_u \subset U \cap U_{i(u)}$ which is standard open in $U_{i(u)}$.
-By compactness we can find finitely many points $u_1, \ldots, u_n \in U$
+$W_u \subset U \cap U_{i(u)}$ containing $u$ which is standard open in $U_{i(u)}$.
+By quasi-compactness we can find finitely many points $u_1, \ldots, u_n \in U$
 such that $U = \bigcup_{j = 1}^n W_{u_j}$. For each $j$ write
 $f^{-1}U_{i(u_j)} = \bigcup_{k \in K_j} V_{jk}$ as a finite
 union of affine opens. Since $W_{u_j} \subset U_{i(u_j)}$ is a standard
@@ -3591,20 +3618,8 @@ \section{Quasi-compact morphisms}
 open of $V_{jk}$, see Algebra, Lemma \ref{algebra-lemma-spec-functorial}.
 Hence $f^{-1}(W_{u_j}) \cap V_{jk}$ is affine, and so
 $f^{-1}(W_{u_j})$ is a finite union of affines. This proves that the
-inverse image of any affine open is a finite union of affine opens.
-
-\medskip\noindent
-Next, assume that the inverse image of every affine open is a finite
-union of affine opens.
-Let $K \subset S$ be any quasi-compact open. Since $S$ has a basis
-of the topology consisting of affine opens we see that $K$ is a finite
-union of affine opens. Hence the inverse image of $K$ is a finite
-union of affine opens. Hence $f$ is quasi-compact.
-
-\medskip\noindent
-Finally, assume that $f$ is quasi-compact. In this case the argument
-of the previous paragraph shows that the inverse image of any affine
-is a finite union of affine opens.
+inverse image of any quasi-compact open is a finite union of affine opens and,
+hence, quasi-compact.
 \end{proof}
 
 \begin{lemma}
@@ -4078,6 +4093,64 @@ \section{Separation axioms}
 which is not quasi-compact.
 \end{example}
 
+\begin{lemma}
+\label{quasi-separated-local-target}
+\begin{slogan}
+(quasi-)separatedness is local on the target.
+\end{slogan}
+For all morphisms of schemes $f:X\to Y$ and all open covers $Y=\bigcup V_i$,
+we have that $f$ is quasi-separated if and only if
+$f|_{f^{-1}(V_i)}:f^{-1}(V_i)\to V_i$ is quasi-separated.
+The same is true if we substitute ``quasi-separated'' by ``separated.''
+\end{lemma}
+
+\begin{proof}
+Let $\mathcal{P}$ be a property of morphisms of schemes stable
+under pre- and postcompositions with isomorphisms
+(i.e., $g\circ f$ and $f\circ h$ have $\mathcal{P}$
+provided that $f$ has $\mathcal{P}$, for all isos $g,h$
+such that the composites make sense).
+We say that a morphism of schemes $f:X\to Y$ has $\Delta_\mathcal{P}$
+if the diagonal map $\Delta_{X/Y}$ has $\mathcal{P}$.
+(The ``isomorphism-stable'' condition on $\mathcal{P}$ is to guarantee
+that $\Delta_\mathcal{P}$ does not depend on the choice of fibre product.)
+We say that $\mathcal{P}$ is {\it local on the target}
+if for all morphisms of schemes $f:X\to Y$ and all open covers $Y=\bigcup V_i$,
+we have that $f$ has $\mathcal{P}$ if and only if
+$f|_{f^{-1}(V_i)}:f^{-1}(V_i)\to V_i$ has $\mathcal{P}$.
+We shall prove that if $\mathcal{P}$ is local on the target,
+then so is $\Delta_\mathcal{P}$.
+The statement will follow then from this result plus
+Lemmas \ref{lemma-quasi-compact-affine} and
+\ref{definition-closed-immersion-locally-ringed-spaces}.
+	
+\medskip\noindent
+Let $f:X\to Y$ be a morphism of schemes.
+
+\medskip\noindent
+Suppose first $f$ has $\mathcal{P}$, let $V\subset Y$ be open and call $U=f^{-1}(V)$.
+Note that $U\times_Y U=U\times_V U$ is an open subscheme of $X\times_Y X$
+(Lemma \ref{lemma-open-fibre-product}). Since $\Delta_f$ has $\mathcal{P}$,
+then so does $\Delta_{X/Y}|_{U}:U=\Delta_{X/Y}^{-1}(U\times_V U)\to U\times_V U$.
+But this map is $\Delta_{U/V}$.
+
+\medskip\noindent
+Conversely, suppose there is an open cover $Y=\bigcup_{i\in I}V_i$
+such that $f|_{U_i}:U_i\to V_i$ has $\Delta_\mathcal{P}$ for all $i\in I$,
+where $U_i=f^{-1}(V_i)$, i.e., $\Delta_{U_i/V_i}$ has $\mathcal{P}$.
+But $\Delta_{U_i/V_i}$ can be identified with
+$\Delta_{X/Y}|_{U_i}:U_i
+=\Delta_{X/Y}^{-1}(U_i\times_{V_i}U_i)
+\to U_i\times_{V_i}U_i\subset X\times_Y X$.
+It suffices to argue then that the sets $U_i\times_{V_i}U_i$ cover $X\times_Y X$.
+This follows from the facts
+(i) $U_i\times_{V_i}U_i=p^{-1}(U_i)\cap q^{-1}(U_i)$,
+where $X\xleftarrow{p}X\times_Y X\xrightarrow{q}X$ are the canonical projections
+(Lemma \ref{lemma-open-fibre-product}),
+(ii) $f\circ p=f\circ q$, and
+(iii) the sets $V_i$ cover $Y$.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-where-are-they-equal}
 Let $X$, $Y$ be schemes over $S$.
@@ -4358,6 +4431,74 @@ \section{Separation axioms}
 that $U \to X$ is separated.
 \end{proof}
 
+Thanks to all the job previously done, in the next two lemmas
+we can give a topological characterization of quasi-separatedness.
+
+\begin{lemma}
+	\label{lemma-characterize-quasi-separated-scheme}
+	Let $X$ be a scheme. The following are equivalent:
+	\begin{enumerate}
+		\item $X$ is quasi-separated,
+		\item The underlying space of $X$ is quasi-separated
+		(Topology, Definition \ref{definition-quasi-separated-space}),
+		\item For all affine open subsets $U,V\subset X$, it holds that $U\cap V$ is quasi-compact,
+		\item There exists an affine open cover $X=\bigcup_{i\in I} U_i$ such that $U_i\cap U_j$ is quasi-compact for all $i,j\in I$, and
+		\item There exists a cover $X=\bigcup_{i\in I}U_i$ of quasi-compact quasi-separated open subschemes such that $U_i\cap U_j$ is quasi-compact for all $i,j\in I$.
+	\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+	(1)$\Rightarrow$(2).
+	Use Lemma \ref{lemma-characterize-quasi-separated} and
+	Topology, Lemma \ref{lemma-quasi-separated-basis}.
+	
+	\medskip\noindent
+	(2)$\Rightarrow$(3).
+	Follows from the fact that affine schemes are quasi-compact,
+	Algebra, Lemma \ref{lemma-quasi-compact}.
+	
+	\medskip\noindent
+	(3)$\Rightarrow$(4). Trivial.
+	
+	\medskip\noindent
+	(4)$\Rightarrow$(5).
+	It follows from the fact that affine schemes are separated
+	(hence, quasi-separated),
+	Lemma \ref{lemma-affine-separated}.
+	
+	\medskip\noindent
+	(5)$\Rightarrow$(1).
+	Since “being quasi-compact” is a property of morphisms of schemes which is local on the target (Lemma \ref{lemma-quasi-compact-affine}), to see that the morphism $\Delta_X:X\to X\times_\mathbb{Z}X$ is quasi-compact it suffices to see that the restriction to the open subset $U_i \times_\mathbb{Z}U_j \subset X\times_\mathbb{Z}X$, which is $\Delta_X^{-1}(U_i \times_\mathbb{Z}U_j )=U_i \cap U_j \to U_i \times_\mathbb{Z}U_j $, is quasi-compact. Since a scheme $Y$ is quasi-compact if and only if $Y\to\Spec \mathbb{Z}$ is quasi-compact (use Lemma \ref{lemma-quasi-compact-affine}), the composite $U_i \cap U_j \to U_i \times_\mathbb{Z}U_j \to\Spec \mathbb{Z}$ is quasi-compact. By Lemma \ref{lemma-quasi-compact-permanence}, it suffices to argue that $U_i \times_\mathbb{Z}U_j \to\Spec \mathbb{Z}$ is quasi-separated. This is because (i) the morphism $U_i \times_\mathbb{Z} U_j \to U_j $ is quasi-separated for it is the base change of the quasi-separated morphism $U_i \to\Spec \mathbb{Z}$ (Lemma \ref{lemma-separated-permanence}), and (ii) quasi-separatedness is stable under compositions (Lemma \ref{lemma-separated-permanence}), so $U_i \times_\mathbb{Z}U_j \to U_j \to\Spec \mathbb{Z}$ is quasi-separated.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-quasi-separatedness-is-topological}
+Let $f:X\to S$ be a morphism of schemes. The following are equivalent:
+\begin{enumerate}
+	\item $f$ is a quasi-separated morphism of schemes,
+	\item $f^{-1}(U)$ is quasi-separated, for all open
+	affine subsets $U\subset S$, and
+	\item the underlying map on topological spaces of $f$ is quasi-separated
+	(see Topology, Definition \ref{definition-quasi-separated-space}).
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+(3)$\Rightarrow$(2).
+This is because affine schemes are quasi-separated, Lemma \ref{lemma-affine-separated}.
+
+\medskip\noindent
+(2)$\Rightarrow$(1).
+Follows by Lemma \ref{lemma-characterize-quasi-separated}
+(and because affine schemes are quasi-separated,
+Lemmas \ref{lemma-affine-separated} and
+\ref{lemma-characterize-quasi-separated-scheme}).
+
+\medskip\noindent
+(1)$\Rightarrow$(3).
+Let $f:X\to Y$ be a quasi-separated morphism of schemes, and let $V\subset Y$ be an open subscheme whose underlying topological space is quasi-separated. Then $f^{-1}(V)\to V$ is quasi-separated (as a morphism of schemes, use for instance Lemma \ref{lemma-characterize-quasi-separated}). By Lemma \ref{lemma-characterize-quasi-separated-scheme}, $V\to\Spec\mathbb{Z}$ is quasi-separated (as a morphism of schemes). Hence, by Lemma \ref{lemma-separated-permanence}, the composite $f^{-1}(V)\to V\to\Spec\mathbb{Z}$ is a quasi-separated morphism of schemes. We finish by Lemma \ref{lemma-characterize-quasi-separated-scheme}.
+\end{proof}
+
 \noindent
 You may have been wondering whether the condition
 of only considering pairs of affine opens whose image is contained
diff --git a/topology.tex b/topology.tex
index 8bffb2c9d..db445105f 100644
--- a/topology.tex
+++ b/topology.tex
@@ -5209,7 +5209,33 @@ \section{Miscellany}
 {\it isolated point} of $X$ if $\{x\}$ is open in $X$.
 \end{definition}
 
+\section{Quasi-separated spaces}
 
+\begin{definition}
+	\label{definition-quasi-separated-space}
+	A topological space $X$ is said to be {\it quasi-separated} if
+	for every open quasi-compact subsets $U,V\subset X$ it holds
+	that $U\cap V$ is quasi-compact. A continuous map of spaces
+	$f:X\to Y$ is {\it quasi-separated} if for every
+	quasi-separated open $V\subset Y$ we have $f^{-1}(V)$ is quasi-separated.
+\end{definition}
+
+\begin{lemma}
+	\label{lemma-quasi-separated-basis}
+	Let $X$ be a space with a basis $\mathcal{B}$ of
+	quasi-compact open sets. Then $X$ is quasi-separated
+	if and only if the intersection of two open sets of
+	$\mathcal{B}$ is quasi-compact.
+\end{lemma}
+
+\begin{proof}
+	The implication to the right is clear. For the converse,
+	let $U,V\subset X$ be quasi-compact open subsets of $X$.
+	Since each of $U$ and $V$ can be written as a finite union
+	of quasi-compact basic open sets, $U\cap V$ can be written
+	as a finite union of intersections of pairs of basic sets,
+	i.e., it is a finite union of quasi-compact sets and, hence, quasi-compact.
+\end{proof}
 
 \section{Partitions and stratifications}
 \label{section-stratifications}

From 79b6ea3909c44f3d796b1ab97e18f3056a10cc6a Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Mon, 12 Jun 2023 18:32:22 +0200
Subject: [PATCH 3/9] Added immersions characterization

Added lemma-immersion, which provides an easy characterization of immersions of schemes.
---
 morphisms.tex | 37 +++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)

diff --git a/morphisms.tex b/morphisms.tex
index 76d88114c..994bad5ba 100644
--- a/morphisms.tex
+++ b/morphisms.tex
@@ -216,6 +216,43 @@ \section{Immersions}
 \noindent
 In this section we collect some facts on immersions.
 
+\begin{lemma}
+\label{lemma-immersion}
+Let $f:X\to Y$ be a morphism of schemes.
+Then $f$ is an immersion if and only if it is a homeomorphism
+onto a locally closed subset of $Y$ and $f^\sharp:f^{-1}\mathcal{O}_Y\to\mathcal{O}_X$
+is surjective.
+\end{lemma}
+
+\begin{proof}
+Suppose $f$ is an immersion. There is an open subset $U\subset Y$
+such that $f$ equals the composite $X\xrightarrow{i}U\xrightarrow{j}Y$,
+where $i$ is a closed immersion. On the level of sheaves, equation $f=j\circ i$
+means that the composite
+$i^{-1}j^{-1}\mathcal{O}_Y
+\xrightarrow{i^{-1}(j^\sharp)}i^{-1}\mathcal{O}_U
+\xrightarrow{i^\sharp}\mathcal{O}_X$ equals $f^\sharp$.
+Since $j^\sharp$ is an isomorphism, we get that $i^\sharp$ is onto.
+Now apply Lemma \ref{lemma-closed-immersion}.
+
+\medskip\noindent
+Conversely, suppose that $f$ is a homeomorphism onto a locally closed
+subset of $Y$ and that $f^{-1}\mathcal{O}_Y\to\mathcal{O}_X$ is surjective.
+There is an open subset $U\subset Y$ such that $f(X)\subset U$ and $f(X)$
+is closed in $U$. By Schemes, Lemma \ref{lemma-restrict-map-to-opens},
+the commutative diagram of topological spaces
+$$
+\xymatrix{
+	X \ar@{->}[r]^{i} \ar@/_/@{->}[rr]_{f} & U \ar@{->}[r]^{j} & Y
+}
+$$
+promotes to a commutative diagram of schemes,
+where the morphism $i^\sharp:i^{-1}\mathcal{O}_U\to\mathcal{O}_X$
+is given by $f^\sharp:f^{-1}\mathcal{O}_Y\cong i^{-1}j^{-1}\mathcal{O}_Y
+\cong i^{-1}\mathcal{O}_U\to\mathcal{O}_X$.
+Apply again Lemma \ref{lemma-closed-immersion}.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-immersion-permanence}
 Let $Z \to Y \to X$ be morphisms of schemes.

From 0647cfa12ddf073b8f045a2e4f4d6947f223fbb7 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Thu, 15 Jun 2023 11:15:57 +0200
Subject: [PATCH 4/9] Added omitted proofs and converses

Changes in:
- schemes.tex: Added proof of the converse of lemma-where-are-they-equal
- morphisms.tex: Given detailed proofs of lemma-integral-local, lemma-finite-local and lemma-base-change-finite
- algebra.tex: Added converse in lemma-integral-local (for it is used in the proof of the homonymous lemma in schemes.tex)
---
 algebra.tex   | 16 +++++++++---
 morphisms.tex | 41 +++++++++++++++++++++++++----
 schemes.tex   | 72 +++++++++++++++++++++++++++++++++++++++++++++------
 3 files changed, 112 insertions(+), 17 deletions(-)

diff --git a/algebra.tex b/algebra.tex
index 6cb53e090..97fd175b3 100644
--- a/algebra.tex
+++ b/algebra.tex
@@ -7589,20 +7589,28 @@ \section{Finite and integral ring extensions}
 Let $R \to S$ be a ring map.
 Let $f_1, \ldots, f_n \in R$ generate the unit ideal.
 \begin{enumerate}
-\item If each $R_{f_i} \to S_{f_i}$ is integral, so is $R \to S$.
-\item If each $R_{f_i} \to S_{f_i}$ is finite, so is $R \to S$.
+\item $R \to S$ is integral if and only if
+$R_{f_i} \to S_{f_i}$ is integral, for all $i=1,\dots,n$.
+\item $R \to S$ is finite if and only if
+$R_{f_i} \to S_{f_i}$ is finite, for all $i=1,\dots,n$.
 \end{enumerate}
 \end{lemma}
 
 \begin{proof}
-Proof of (1).
+Suppose that $R\to S$ is integral (resp., finite).
+By Lemma \ref{lemma-base-change-integral},
+each $R_{f_i} \to S_{f_i}$ is integral
+(resp., finite), using Lemma \ref{lemma-tensor-localization}).
+The converse of part (2) is a particular case of Lemma \ref{lemma-cover}.
+We prove the converse of part (1).
+Suppose then $R_{f_i} \to S_{f_i}$ is integral, for all $i=1,\dots,n$.
 Let $s \in S$. Consider the ideal $I \subset R[x]$ of
 polynomials $P$ such that $P(s) = 0$. Let $J \subset R$
 denote the ideal (!) of leading coefficients of elements of $I$.
 By assumption and clearing denominators
 we see that $f_i^{n_i} \in J$ for all $i$
 and certain $n_i \geq 0$. Hence $J$ contains $1$ and we see
-$s$ is integral over $R$. Proof of (2) omitted.
+$s$ is integral over $R$.
 \end{proof}
 
 \begin{lemma}
diff --git a/morphisms.tex b/morphisms.tex
index 994bad5ba..1a6a32ffb 100644
--- a/morphisms.tex
+++ b/morphisms.tex
@@ -10700,8 +10700,16 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-integral-local}.
-Some details omitted.
+Implications (1)$\Rightarrow$(3)$\Rightarrow$(2) are clear.
+We see (2)$\Rightarrow$(1).
+Let $V=\Spec A\subset S$ be open affine,
+By Lemma \ref{lemma-characterize-affine}, $f^{-1}(V)=\Spec B$ is affine.
+Since any localization of an integral ring morphism is again integral
+(Algebra, Lemma \ref{algebra-lemma-integral-local}), we may
+(using Schemes, Lemma \ref{lemma-standard-open-two-affines})
+cover $\Spec A$ by affine open subschemes of the form $D(f_i)$,
+$f_i\in A$, $i=1,\dots,n$, such that $B_{f_i}$ is an integral $A_{f_i}$-algebra.
+By Algebra, Lemma \ref{algebra-lemma-integral-local}, $A\to B$ is integral.
 \end{proof}
 
 \begin{lemma}
@@ -10721,8 +10729,8 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-integral-local}.
-Some details omitted.
+The proof is exactly the same as the proof of Lemma \ref{lemma-integral-local},
+but substituting ``integral'' by ``finite.''
 \end{proof}
 
 \begin{lemma}
@@ -10754,7 +10762,30 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-base-change-integral}.
+Let
+$$
+\xymatrix{
+	X' \ar@{->}[r]^{f'} \ar@{->}[d] & S' \ar@{->}[d] \\
+	X \ar@{->}[r]^{f} & S
+}
+$$
+be a cartesian diagram of schemes, and suppose that $f$ is finite
+(resp., integral). Lemma \ref{lemma-base-change-affine} tells us that
+$f'$ is affine. Each point $s'\in S$ has an affine open neighborhood
+$V\subset S'$ that maps to an open affine $U\subset S$.
+In the proof of the last invoked lemma one sees that the restricted diagram
+$$
+\xymatrix{
+	(f')^{-1}(V) \ar@{->}[d] \ar@{->}[r]^{f'} & V \ar@{->}[d] \\
+	f^{-1}(U) \ar@{->}[r]^{f} & U
+}
+$$
+is a cartesian square of affine schemes.
+Hence, using Algebra, Lemma \ref{algebra-lemma-base-change-integral}, one sees that
+$f'$ is finite, by Lemma \ref{lemma-finite-local}
+(resp., that $f'$ is integral, by Lemma \ref{lemma-integral-local}).
+Since these $V$ cover $S'$, we finish by Lemma \ref{lemma-finite-local}
+(resp., by Lemma \ref{lemma-integral-local}).
 \end{proof}
 
 \begin{lemma}
diff --git a/schemes.tex b/schemes.tex
index c37ac483b..42ecad047 100644
--- a/schemes.tex
+++ b/schemes.tex
@@ -4151,18 +4151,33 @@ \section{Separation axioms}
 (iii) the sets $V_i$ cover $Y$.
 \end{proof}
 
+A topological space $Y$ is Hausdorff if for all topological spaces $X$
+and all morphisms $a,b:X\to Y$, the equalizer of $(a,b)$ is a closed subset of $X$.
+In schemes we have the following result:
+
 \begin{lemma}
 \label{lemma-where-are-they-equal}
-Let $X$, $Y$ be schemes over $S$.
-Let $a, b : X \to Y$ be morphisms of schemes over $S$.
+Let $Y$ be a scheme over $S$.
+\begin{enumerate}
+\item Let $X$ be a scheme over $S$ and let
+$a, b : X \to Y$ be morphisms of schemes over $S$.
 There exists a largest locally closed subscheme
 $Z \subset X$ such that $a|_Z = b|_Z$. In fact $Z$ is
-the equalizer of $(a, b)$. Moreover, if $Y$ is separated
-over $S$, then $Z$ is a closed subscheme.
+the equalizer of $(a, b)$ in both $Sch/S$ and $Sch$.
+In particular, all equalizers exist in $Sch/S$ and $Sch$.
+\item The following are equivalent:
+\begin{enumerate}
+\item $Y$ is separated over $S$,
+\item for all affine schemes $X$ over $S$ and all morphisms $a,b:X\to Y$
+over $S$, the equalizer of $(a,b)$ is a closed subscheme of $X$, and
+\item for all schemes $X$ over $S$ and all morphisms $a,b:X\to Y$
+over $S$, the equalizer of $(a,b)$ is a closed subscheme of $X$.
+\end{enumerate}
+\end{enumerate}
 \end{lemma}
 
 \begin{proof}
-The equalizer of $(a, b)$ is for categorical reasons
+The equalizer of $(a, b)$ in $Sch/S$ is for categorical reasons
 the fibre product $Z$ in the following diagram
 $$
 \xymatrix{
@@ -4171,11 +4186,52 @@ \section{Separation axioms}
 Y \ar[r]^-{\Delta_{Y/S}} & Y \times_S Y
 }
 $$
-Thus the lemma follows from Lemmas
-\ref{lemma-base-change-immersion}, \ref{lemma-diagonal-immersion} and
-Definition \ref{definition-separated}.
+Thus part (1) follows from Lemmas
+\ref{lemma-base-change-immersion} and \ref{lemma-diagonal-immersion}.
+On the other hand, if $\mathcal{C}$ is any category and $c\in\text{Ob}(\mathcal{C})$,
+then it is easy to see that the forgetful functor $\mathcal{C}/c\to\mathcal{C}$
+preserves equalizers. It follows that $Z$ is also the equalizer in $Sch$.
+The last assertion comes from $Sch/\Spec\mathbb{Z}=Sch$.
+This finishes the proof of (1). We now see part (2).
+
+\medskip\noindent
+(a)$\Rightarrow$(c).
+It follows from the usage of the cartesian diagram above plus the same mentioned lemmas.
+
+\medskip\noindent
+(c)$\Rightarrow$(b). Trivial.
+
+\medskip\noindent
+(b)$\Rightarrow$(c).
+Let be $a,b:X\to Y$ morphisms of schemes and let $i:Z\to X$ be their equalizer.
+By part (1), $i$ is an immersion. By Lemma \ref{lemma-immersion-when-closed},
+it suffices to see that $i(Z)$ is closed in $X$. Equivalently, $i(Z)\cap U$
+is closed in $U$, for $U\subset X$ open affine. In turn, by what (b) says,
+it suffices to show that $i^{-1}(U)\to U$ is the equalizer of $(a|_U,b|_U)$.
+Let $K\to U$ be the equalizer of $(a|_U,b|_U)$ and let $K\to Z$ be the
+canonical morphism. If we show that the diagram
+$$
+\xymatrix{
+K \ar@{->}[r] \ar@{->}[d] & U \ar@{->}[d] \\
+Z \ar@{->}[r]^i & X
+}
+$$
+is cartesian, it will follow that $K\cong i^{-1}(U)$.
+Verifying that $Z\leftarrow K\to U$ satisfies the cartesianity
+condition is easy (use Lemma \ref{lemma-immersions-monomorphisms}).
+
+\medskip\noindent
+(c)$\Rightarrow$(a).
+The diagonal morphism of $Y\to S$ is the equalizer of the two
+projections $Y\times_SY\to Y$ (true in any category with fibre products).
 \end{proof}
 
+\begin{remark}
+\label{remark-where-are-they-equal}
+A characterization of separated schemes is obtained from erasing all
+``over $S$'' in part (2) of Lemma \ref{lemma-where-are-they-equal}.
+\end{remark}
+
 \begin{lemma}
 \label{lemma-characterize-quasi-separated}
 Let $f : X \to S$ be a morphism of schemes.

From c5edb6794f370568ac8ecaf4cef275399fac241d Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Sun, 18 Jun 2023 17:07:23 +0200
Subject: [PATCH 5/9] Added new results that were being implicitly used
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

Changes in:

- topology.tex:
· added the new lemma-correspondence-sober-generic-points-open. I've noticed that this result is used implicitly in the sections "Relative normalization" and "Normalization" of the Morphisms chapter. It took me a while to find out about it, so I thought it's worth to be added.
· Added the new lemma-correspondence-closed-irreducible-open, used in the proof of the former.

- schemes.tex: Added a new result, lemma-quasi-separated-coproduct, (and the auxiliary lemma-fibre-product-commutes-coproduct) that was implicitly used in morphisms.tex to show that 035M is quasi-separated (it's now invoked there).

- morphisms.tex:
· given more detail in the proof of lemma-normalization-generic with the aid of the afore-mentioned newly added result in topology.tex
· Given more detail in the final step of the proof of lemma-description-normalization.
---
 morphisms.tex | 23 +++++++++++++++++++----
 schemes.tex   | 52 ++++++++++++++++++++++++++++++++++++++++++++++++++-
 topology.tex  | 49 ++++++++++++++++++++++++++++++++++++++++++++++++
 3 files changed, 119 insertions(+), 5 deletions(-)

diff --git a/morphisms.tex b/morphisms.tex
index 1a6a32ffb..2ee941672 100644
--- a/morphisms.tex
+++ b/morphisms.tex
@@ -13756,12 +13756,21 @@ \section{Relative normalization}
 \end{lemma}
 
 \begin{proof}
-By Lemma \ref{lemma-normalization-localization} we may assume $X = \Spec(A)$
+For continuous maps of sober spaces, the following property
+is local on the target: ``all generic points of the irreducible
+components of the target are hit by generic points of irreducible
+components of the source'' (use Topology, Lemma
+\ref{topology-lemma-correspondence-sober-generic-points-open}).
+Hence, by Lemma \ref{lemma-normalization-localization} we may assume $X = \Spec(A)$
 is affine. Choose a finite affine open covering $Y = \bigcup \Spec(B_i)$.
 Then $X' = \Spec(A')$ and the morphisms $\Spec(B_i) \to Y \to X'$
 jointly define an injective $A$-algebra map $A' \to \prod B_i$.
 Thus the lemma follows from
-Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}.
+Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}
+(and also that by Topology, Lemma
+\ref{topology-lemma-correspondence-sober-generic-points-open},
+every generic point of an irreducible component in $Y$ is a generic
+point in $\Spec B_i$ of an irreducible component, for some $i$).
 \end{proof}
 
 \begin{lemma}
@@ -14045,7 +14054,9 @@ \section{Normalization}
 canonical maps of Schemes, Section \ref{schemes-section-points}.
 Note that this morphism is quasi-compact by assumption and
 quasi-separated as $Y$ is separated (see
-Schemes, Section \ref{schemes-section-separation-axioms}).
+Schemes, Lemmas \ref{lemma-compose-after-separated},
+\ref{schemes-lemma-quasi-separated-coproduct}, and
+\ref{schemes-lemma-affine-separated}).
 
 \begin{definition}
 \label{definition-normalization}
@@ -14125,13 +14136,17 @@ \section{Normalization}
 This proves (2). Part (3) follows from
 Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal},
 or Lemma \ref{lemma-normalization-in-disjoint-union}.
-Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism
+Finally, we prove part (4).
+By Lemma \ref{lemma-normalization-localization} and Topology,
+Lemma \ref{topology-lemma-correspondence-sober-generic-points-open},
+the morphism $f^{-1}(U) \to U$ is the morphism
 $$
 \Spec\left(\prod \kappa(\mathfrak q_i)\right)
 \longrightarrow
 \Spec(A)
 $$
 where $f : Y \to X$ is the morphism (\ref{equation-generic-points}).
+Hence (4) follows from (2) and (3).
 \end{proof}
 
 \begin{lemma}
diff --git a/schemes.tex b/schemes.tex
index 42ecad047..ee1c8b4cd 100644
--- a/schemes.tex
+++ b/schemes.tex
@@ -3328,6 +3328,43 @@ \section{Fibre products of schemes}
 to each other.
 \end{proof}
 
+\begin{lemma}
+\label{lemma-fibre-product-commutes-coproduct}
+\begin{slogan}
+	Fibre products of schemes commute with coproducts.
+\end{slogan}
+Let $S$ be a scheme, and let $X_i$, $Y_j$,
+$i\in I$, $j\in J$ be schemes over $S$. Then
+$$
+\left(\coprod_i X_i\right)
+\times_S\left(\coprod_j Y_j\right)
+=\coprod_{ij} X_i\times_S Y_j.
+$$
+\end{lemma}
+
+\begin{proof}
+Let $Z$ be a scheme, and let $f:Z\to\coprod_i X_i$,
+$g:Z\to\coprod_j Y_j$ be morphisms of schemes such that the diagram
+$$
+\xymatrix{
+	Z \ar@{->}[r]^{f} \ar@{->}[d]_{g} & \coprod_i X_i \ar@{->}[d] \\
+	\coprod_j Y_j \ar@{->}[r] & S
+}
+$$
+commutes.
+Note that we have canonical morphisms from
+$\coprod_{ij} X_i\times_S Y_j$ into $\coprod X_i$
+and into $\coprod Y_j$, respectively.
+Let $U_{ij}=f^{-1}(X_i)\cap g^{-1}(Y_j)$.
+Then the sets $U_{ij}$ are open, disjoint and cover $Z$,
+i.e., $Z=\coprod_{ij}U_{ij}$. By commutativity of the last diagram,
+there is an induced morphism $U_{ij}\to X_i\times _S Y_j$.
+Hence, since $Z$ is the coproduct of the schemes $U_{ij}$,
+these maps assemble into a morphism $Z\to \coprod_{ij} X_i\times_S Y_j$,
+through which $f$ and $g$ factor. One can also verify
+that such a morphism is uniquely determined by $f$ and $g$.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-fibre-product-immersion}
 Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes
@@ -4093,6 +4130,19 @@ \section{Separation axioms}
 which is not quasi-compact.
 \end{example}
 
+\begin{lemma}
+	\label{lemma-quasi-separated-coproduct}
+	Let $f_i:X_i\to S$ be a morphism of schemes, $i\in I$.
+	For all $i\in I$, suppose $f_i$ is quasi-separated (resp., separated).
+	Then $\coprod f_i:\coprod X_i\to S$ is quasi-separated (resp., separated).
+\end{lemma}
+
+\begin{proof}
+	By Lemma \ref{lemma-fibre-product-commutes-coproduct}, the diagonal morphism of $\coprod f_i$ is the unique map $\coprod X_i\to\coprod X_i\times_S X_j$ such that precomposed with the inclusion $X_k\to\coprod X_i$ equals $X_k\xrightarrow{\Delta_{X_k/S}} X_k\times_S X_k\to \coprod X_i\times_S X_j$. That is, $\Delta_{\coprod f_i}^{-1}(X_i\times_S X_j)$ equals the empty scheme if $i\neq j$ and equals $X_i\times_S X_i$ if $i=j$.
+	By Lemmas \ref{lemma-quasi-compact-affine} and
+	\ref{lemma-closed-local-target}, the result follows.
+\end{proof}
+
 \begin{lemma}
 \label{quasi-separated-local-target}
 \begin{slogan}
@@ -4122,7 +4172,7 @@ \section{Separation axioms}
 then so is $\Delta_\mathcal{P}$.
 The statement will follow then from this result plus
 Lemmas \ref{lemma-quasi-compact-affine} and
-\ref{definition-closed-immersion-locally-ringed-spaces}.
+\ref{lemma-closed-local-target}.
 	
 \medskip\noindent
 Let $f:X\to Y$ be a morphism of schemes.
diff --git a/topology.tex b/topology.tex
index db445105f..42bfac507 100644
--- a/topology.tex
+++ b/topology.tex
@@ -765,6 +765,34 @@ \section{Irreducible components}
 as a singleton space is irreducible.
 \end{proof}
 
+\begin{lemma}
+\label{lemma-correspondence-closed-irreducible-open}
+Let $X$ be a topological space. There is a one-to-one correspondence
+$$
+\begin{Bmatrix}
+\text{Irreducible closed}\\
+\text{subsets of } X \text{ meeting } U
+\end{Bmatrix}
+\longleftrightarrow
+\begin{Bmatrix}
+\text{Irreducible closed}\\
+\text{subsets of } U
+\end{Bmatrix}
+$$
+Moreover, the assignments restrict to a correspondence
+between the set of irreducible components of $X$ meeting $U$
+and the set of irreducible components of $U$.
+\end{lemma}
+
+\begin{proof}
+The map to the right is $Z\mapsto Z\cap U$,
+whereas the map to the left is $F\mapsto \overline{F}$.
+We omit the verification that the maps are mutually inverse.
+On the other hand, these maps preserve inclusions, so we actually
+have a poset isomorphism. In particular, it preserves maximal elements,
+i.e., the irreducible components.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-pick-irreducible-components}
 Let $X$ be a topological space and suppose $X = \bigcup_{i = 1, \ldots, n} X_i$
@@ -915,6 +943,27 @@ \section{Irreducible components}
 Proof of (3). Immediately from (1) and (2).
 \end{proof}
 
+\begin{lemma}
+\label{lemma-correspondence-sober-generic-points-open}
+Let $X$ be a topological space, $\xi\in X$ be a point and $U\subset X$
+be an open subset. Then $\xi$ is a generic point in $X$ of an
+irreducible subset meeting $U$ if and only if $\xi$ is a generic point in $U$.
+In particular, if $X$ is sober, then so is $U$ and the generic
+points of the irreducible components of $X$ that meet $U$ are
+exactly the generic points of the irreducible components of $U$.
+\end{lemma}
+\begin{proof}
+The first assertion can be written symbolically as
+``$\text{cl}_X(\{\xi\})$ is irreducible and meets $U$
+if and only if $\xi\in U$ and $\text{cl}_U(\{\xi\})$ is irreducible.''
+We leave the proof to the reader. In particular, if the conditions hold,
+you can verify that $\text{cl}_X(\{\xi\})$ corresponds to
+$\text{cl}_U(\{\xi\})$ in the bijection of
+Lemma \ref{lemma-correspondence-closed-irreducible-open}.
+Hence, the second assertion follows then from this lemma
+($U$ is sober by Lemma \ref{lemma-sober-local}).
+\end{proof}
+
 \begin{example}
 \label{example-quasi-sober-not-kolmogorov}
 Let $X$ be an indiscrete space of cardinality at least $2$. Then $X$ is

From fd108cb1c502000d19a1c0e415857e0151a0bcc3 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Sun, 18 Jun 2023 17:15:19 +0200
Subject: [PATCH 6/9] Amended coding style errors

---
 categories.tex | 54 +++++++++++++++++++++++++-------------------------
 schemes.tex    |  6 +++---
 2 files changed, 30 insertions(+), 30 deletions(-)

diff --git a/categories.tex b/categories.tex
index 755e11973..a43761537 100644
--- a/categories.tex
+++ b/categories.tex
@@ -850,33 +850,33 @@ \section{Fibre products}
 \end{proof}
 
 \begin{lemma}
-	\label{lemma-pasting-law-pullbacks}
-	Let
-	$$
-	\xymatrix{
-		X' \ar@{->}[r] \ar@{->}[d] & X \ar@{->}[d] \\
-		Y' \ar@{->}[r] \ar@{->}[d] & T \ar@{->}[d] \\
-		Y \ar@{->}[r] & S
-	}
-	$$
-	be a commutative diagram in a category,
-	and suppose that the lower square is cartesian.
-	Then the outer rectangle is cartesian
-	if and only if the upper square is cartesian.
-\end{lemma}
-
-\begin{proof}
-	Suppose each inner square is cartesian. Then the outer rectangle is cartesian by Lemma \ref{lemma-3x3-pullback} applied to the diagram
-	$$
-	\xymatrix{
-		X \ar[r]\ar[d] & T \ar[d] & T \ar[l]\ar[d]\\
-		S \ar[r]       & S        & S \ar[l] \\
-		S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
-	}
-	$$
-	
-	\medskip\noindent
-	We omit the proof of the other implication.
+\label{lemma-pasting-law-pullbacks}
+Let
+$$
+\xymatrix{
+	X' \ar@{->}[r] \ar@{->}[d] & X \ar@{->}[d] \\
+	Y' \ar@{->}[r] \ar@{->}[d] & T \ar@{->}[d] \\
+	Y \ar@{->}[r] & S
+}
+$$
+be a commutative diagram in a category,
+and suppose that the lower square is cartesian.
+Then the outer rectangle is cartesian
+if and only if the upper square is cartesian.
+\end{lemma}
+
+\begin{proof}
+Suppose each inner square is cartesian. Then the outer rectangle is cartesian by Lemma \ref{lemma-3x3-pullback} applied to the diagram
+$$
+\xymatrix{
+	X \ar[r]\ar[d] & T \ar[d] & T \ar[l]\ar[d]\\
+	S \ar[r]       & S        & S \ar[l] \\
+	S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
+}
+$$
+
+\medskip\noindent
+We omit the proof of the other implication.
 \end{proof}
 
 \begin{definition}
diff --git a/schemes.tex b/schemes.tex
index ee1c8b4cd..ac5f0807c 100644
--- a/schemes.tex
+++ b/schemes.tex
@@ -3512,7 +3512,7 @@ \section{Base change in algebraic geometry}
 }
 $$
 where the top square is cartesian by
-Categories, Lemma \ref{lemma-pasting-law-pullbacks}.
+Categories, Lemma \ref{categories-lemma-pasting-law-pullbacks}.
 To see ($\Leftarrow$), use again the same diagram
 but with $Y=S$, so $Y_{S'}=S'$.
 \end{remark}
@@ -4407,7 +4407,7 @@ \section{Separation axioms}
 
 \begin{proof}
 Use the cartesianity of the square from Categories,
-Lemma \ref{lemma-magic-square} and apply
+Lemma \ref{categories-lemma-magic-square} and apply
 Lemmas \ref{lemma-diagonal-immersion},
 \ref{lemma-fibre-product-immersion} and
 \ref{lemma-quasi-compact-preserved-base-change}.
@@ -4465,7 +4465,7 @@ \section{Separation axioms}
 
 \medskip\noindent
 (3) and (4) follow from Categories,
-Lemma \ref{lemma-base-change-diagonal-1-cat} plus
+Lemma \ref{categories-lemma-base-change-diagonal-1-cat} plus
 Lemmas \ref{lemma-fibre-product-immersion} and
 \ref{lemma-quasi-compact-preserved-base-change}.
 (5) and (6) are a particular case of (3) and (4).

From 3fa0d178d385e6292d50c4c65d8561d1af62fabc Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Fri, 23 Jun 2023 13:49:19 +0200
Subject: [PATCH 7/9] Added new section on schemes with locally finitely many
 irreducible components

I regarded the characterization interesting for these schemes are used (among other places) in morphisms.tex, section-normalization and divisors.tex, section-meromorphic-functions and section-meromorphic-reduced
---
 properties.tex | 33 +++++++++++++++++++++++++++++++++
 1 file changed, 33 insertions(+)

diff --git a/properties.tex b/properties.tex
index d38ed6bef..2f59e1652 100644
--- a/properties.tex
+++ b/properties.tex
@@ -731,7 +731,40 @@ \section{Noetherian schemes}
 
 
 
+\section{Schemes with locally finitely many irreducible components}
+\label{section-locally-finitely-many-irred-comp}
 
+\begin{lemma}
+	\label{lemma-characterize-locally-finitely-many-irred-comp}
+	Let $X$ be a scheme. The following are equivalent:
+	\begin{enumerate}
+		\item Every point of $X$ has an open neighborhood that
+		has finitely many irreducible components,
+		\item every quasi-compact open of $X$ has finitely many
+		irreducible components, and
+		\item every affine open of $X$ has finitely many irreducible components.
+	\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+	Implications (2)$\Rightarrow$(3)$\Rightarrow$(1) are trivial, so we only prove
+	(1)$\Rightarrow$(2).
+	Let $U\subset X$ be quasi-compact. There are open subsets
+	$V_1,\dots, V_n\subset X$, each with finitely many irreducible components,
+	such that $U\subset\bigcup_{i=1}^n V_i$. 
+	Thus, $U$ has finitely many irreducible components by
+	Topology, Lemma
+	\ref{topology-lemma-correspondence-sober-generic-points-open}.
+\end{proof}
+
+\begin{definition}
+	\label{definition-locally-finitely-many-irred-comp}
+	Let $X$ be a scheme, if any of the equivalent conditions of
+	Lemma \ref{lemma-characterize-locally-finitely-many-irred-comp} holds,
+	we say that $X$ {\it has locally finitely many irreducible components}.
+\end{definition}
+
+For instance, locally Noetherian schemes have locally finitely many irreducible components.
 
 
 

From 143c0f8de269ab5679601cf5ce7c98133fc5f2e2 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Sun, 25 Jun 2023 13:20:31 +0200
Subject: [PATCH 8/9] Added definition of a previously undefined notion

The purpose of this commit is to add the previously omitted argument in the last sentence of the proof of lemma-normal-normalization (Chapter on morphisms). For this, I defined in the topology chapter what does it mean for a continuous map of spaces to "induce a bijection on irreducible components." This notion is used undefined for instance in 037A, and in lemma-normal-normalization itself (although its definition is somehow acknowledged in https://stacks.math.columbia.edu/tag/04KX#comment-7268). I also added the obligatory characterization of the notion for sober spaces, lemma-characterize-bijection-irreducible-components-sober.

(We also add the reference to "by a topological argument which we omit" in the proof of lemma-normal-normalization.)
---
 morphisms.tex  |  18 +++++---
 properties.tex |   4 +-
 topology.tex   | 113 +++++++++++++++++++++++++++++++++++++++++++++++++
 3 files changed, 127 insertions(+), 8 deletions(-)

diff --git a/morphisms.tex b/morphisms.tex
index 2ee941672..ff0b213a3 100644
--- a/morphisms.tex
+++ b/morphisms.tex
@@ -13853,7 +13853,9 @@ \section{Relative normalization}
 of schemes. Let $X'$ be the normalization of $X$ in $Y$. Assume
 \begin{enumerate}
 \item $Y$ is a normal scheme,
-\item quasi-compact opens of $Y$ have finitely many irreducible components.
+\item $Y$ has locally finitely many irreducible components
+(see Properties, Section
+\ref{properties-section-locally-finitely-many-irred-comp}).
 \end{enumerate}
 Then $X'$ is a disjoint union of integral normal schemes. Moreover,
 the morphism $Y \to X'$ is dominant and induces a bijection of
@@ -13888,14 +13890,18 @@ \section{Relative normalization}
 Properties, Lemma \ref{properties-lemma-locally-normal} that $X'$ is normal.
 On the other hand, each $U'$ is a finite disjoint union of irreducible
 schemes, hence every quasi-compact open of $X'$ has finitely many irreducible
-components (by a topological argument which we omit). Thus $X'$
+components (Properties, Lemma
+\ref{properties-lemma-characterize-locally-finitely-many-irred-comp}).
+Thus $X'$
 is a disjoint union of normal integral schemes by
 Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}.
-It is clear from the description of $X'$ above that $Y \to X'$
-is dominant and induces a bijection on irreducible components
-$V \to U'$ for every affine open $U \subset X$. The bijection of
+It is clear that $V \to U'$ induces a bijection on irreducible components
+for every affine open $U \subset X$. The bijection of
 irreducible components for the morphism $Y \to X'$
-follows from this by a topological argument (omitted).
+follows from this plus Topology, Lemma
+\ref{topology-lemma-bijection-irreducible-components-local-target}.
+The morphism $Y\to X'$ is dominant by Lemma
+\ref{lemma-characterize-normalization}.
 \end{proof}
 
 \begin{lemma}
diff --git a/properties.tex b/properties.tex
index 2f59e1652..d6f24c407 100644
--- a/properties.tex
+++ b/properties.tex
@@ -754,12 +754,12 @@ \section{Schemes with locally finitely many irreducible components}
 	such that $U\subset\bigcup_{i=1}^n V_i$. 
 	Thus, $U$ has finitely many irreducible components by
 	Topology, Lemma
-	\ref{topology-lemma-correspondence-sober-generic-points-open}.
+	\ref{topology-lemma-correspondence-closed-irreducible-open}.
 \end{proof}
 
 \begin{definition}
 	\label{definition-locally-finitely-many-irred-comp}
-	Let $X$ be a scheme, if any of the equivalent conditions of
+	Let $X$ be a scheme. If any of the equivalent conditions of
 	Lemma \ref{lemma-characterize-locally-finitely-many-irred-comp} holds,
 	we say that $X$ {\it has locally finitely many irreducible components}.
 \end{definition}
diff --git a/topology.tex b/topology.tex
index 42bfac507..d949d14eb 100644
--- a/topology.tex
+++ b/topology.tex
@@ -1028,6 +1028,87 @@ \section{Irreducible components}
 a contradiction.
 \end{proof}
 
+\begin{definition}
+\label{definition-bijection-irreducible-components}
+We say that a continuous map $f:X\to Y$ of topological spaces
+{\it induces a bijection between irreducible components}
+if for all irreducible components $Z\subset X$, it holds that
+$\overline{f(Z)}$ is an irreducible component and the map
+$Z\mapsto \overline{f(Z)}$ is a bijection between sets of irreducible components.
+\end{definition}
+
+\begin{lemma}
+\label{lemma-bijection-irreducible-components-local-target}
+Let $f:X\to Y$ be a continuous map of topological spaces.
+For all open coverings $Y=\bigcup V_i$, we have that $f$
+induces a bijection between irreducible components if and only if,
+for all $i$, $f^{-1}(V_i)\to V_i$ induces a bijection between
+irreducible components.
+\end{lemma}
+
+\begin{proof}
+For this proof, we want to note that if a function between topological spaces
+$f:X\to Y$ is continuous then
+$\overline{f(A)}=\overline{f(\overline{A})}$, for all subsets $A\subset X$.
+We will use this property without further mention.
+
+\medskip\noindent
+Suppose first that $f:X\to Y$ induces a bijection on irreducible components and let
+$V\subset Y$ be an open subset.
+We are going to show that $f|_{f^{-1}(V)}:f^{-1}(V)\to V$
+induces a bijection on irreducible components. Let $F\subset f^{-1}(V)$
+be an irreducible component of $f^{-1}(V)$. Then $\overline{F}$ is an
+irreducible component of $X$
+(Lemma \ref{lemma-correspondence-closed-irreducible-open}). Hence,
+$\overline{f(\overline{F})}$ is an irreducible component of $Y$.
+Thus, $\overline{f(\overline{F})}\cap V=\overline{f(F)}\cap V$,
+the closure of $f(F)$ in $V$, is an irreducible component of $V$
+(by same lemma). Now suppose $F,G\subset f^{-1}(V)$ are two
+irreducible components of $f^{-1}(V)$ such that
+$\overline{f(F)}\cap V=\overline{f(G)}\cap V$, i.e.,
+$\overline{f(\overline{F})}\cap V
+=\overline{f(\overline{G})}\cap V$.
+Hence, $\overline{f(\overline{F})}
+=\overline{f(\overline{G})}$ (since $\overline{f(\overline{F})}$
+and $\overline{f(\overline{G})}$ are closed irreducible subsets of $Y$,
+plus last invoked lemma). Thus, $\overline{F}=\overline{G}$, whence $F=G$.
+Now suppose that $T\subset V$ is an irreducible component in $V$.
+Then $\overline{T}$ is an irreducible component in $Y$, so there is
+an irreducible component $F\subset X$ such that
+$\overline{f(F)}=\overline{T}$.
+Hence, $\overline{f(F)}\cap V=T\neq\emptyset$.
+Since $V$ is open, $f(F)\cap V$ is non-empty, so $F\cap f^{-1}(V)$
+is non-empty. In other words, this last set is an irreducible component
+in $f^{-1}(V)$. We also have
+$\overline{f(F\cap f^{-1}(V))}\cap V=\overline{f(F)}\cap V=T$.
+This finishes the proof that $f|_{f^{-1}(V)}:f^{-1}(V)\to V$
+induces a bijection on irreducible components.
+
+\medskip\noindent
+Conversely, suppose $Y=\bigcup V_i$ is an open cover such that
+$f^{-1}(V_i)\to V_i$ induces a bijection on irreducible components, for all $i$.
+Let $F\subset X$ be an irreducible component and let $i$ be such that
+$f^{-1}(V_i)$ meets $F$.
+Then $\overline{f(F\cap f^{-1}(V_i))}\cap V_i
+=\overline{f(F)}\cap V_i$ is an irreducible component in $V_i$,
+whence $\overline{f(F)}$ is an irreducible component in $Y$.
+Suppose $F,G\subset X$ are irreducible components such that
+$\overline{f(F)}=\overline{f(G)}$, and let $i$ be such that
+$V_i$ meets this set. Then $V_i$ meets both $f(F)$ and $f(G)$,
+so $f^{-1}(V_i)$ meets $F$ and $G$. Since
+$\overline{f(F\cap f^{-1}(V_i))}\cap V_i
+=\overline{f(F)}\cap V_i=\overline{f(G)}\cap V_i
+=\overline{f(G\cap f^{-1}(V_i))}\cap V_i$,
+we deduce $F\cap f^{-1}(V_i)=G\cap f^{-1}(V_i)$, whence $F=G$.
+Finally, suppose $T\subset Y$ is an irreducible component,
+and let $i$ be such that $V_i$ meets $T$. Let $F\subset f^{-1}(V_i)$
+be the irreducible component of $f^{-1}(V_i)$ such that
+$\overline{f(\overline{F})}\cap V_i
+=\overline{f(F)}\cap V_i
+=T\cap V_i$.
+This implies $\overline{f(\overline{F})}=T$.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-irreducible-fibres-irreducible-components}
 Let $f : X \to Y$ be a continuous map of topological spaces.
@@ -1109,6 +1190,38 @@ \section{Irreducible components}
 Kolmogorov spaces.
 \end{proof}
 
+\begin{lemma}
+\label{lemma-point-image-map-sober}
+Let $f:X\to Y$ be a continuous map of sober spaces, and let $x\in X$, $y\in Y$
+be points. Then $f(x)=y$ if and only if
+$\overline{f\left(\overline{\{x\}}\right)}=\overline{\{y\}}$.
+\end{lemma}
+
+\begin{proof}
+Denote $c:X\to X'$ to the map constructed in the proof of
+Lemma \ref{lemma-make-sober}, and denote $d:Y\to Y'$ to the analogous map for $Y$.
+Then $c$ and $d$ are homeomorphisms (the identities $\text{id}_X$, $\text{id}_Y$
+are soberifications of $X$ and $Y$). If we denote $f':X'\to Y'$ to the unique
+continuous map such that $f'\circ c=d\circ f$, then the result follows from
+interpreting the condition $f(x)=y$ for $f'$.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-characterize-bijection-irreducible-components-sober}
+Let $f:X\to Y$ be a continuous map between sober topological spaces.
+Then $f$ induces a bijection between irreducible components
+(see Definition \ref{definition-bijection-irreducible-components})
+if and only if
+for every generic point $\eta\in X$ of an irreducible component of $X$,
+$f(\eta)$ is a generic point of an irreducible component of $Y$ and
+$\eta\mapsto f(\eta)$ is a bijection between sets of generic points of
+irreducible components.
+\end{lemma}
+
+\begin{proof}
+It follows from Lemma \ref{lemma-point-image-map-sober}.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-remove-irreducible-connected}
 Let $X$ be a connected topological space with a finite number of

From f6a176a04c71e2293de8f08ef0ae5ddb17d40571 Mon Sep 17 00:00:00 2001
From: eliasgv3 <eliasgv3@gmail.com>
Date: Mon, 26 Jun 2023 12:16:38 +0200
Subject: [PATCH 9/9] Added detail in proof of a lemma

Added detail in a step of a proof of lemma-normalization-normal in the Morphisms chapter.
---
 morphisms.tex |  8 +++++++-
 topology.tex  | 24 ++++++++++++++++++++++++
 2 files changed, 31 insertions(+), 1 deletion(-)

diff --git a/morphisms.tex b/morphisms.tex
index ff0b213a3..20ff244e4 100644
--- a/morphisms.tex
+++ b/morphisms.tex
@@ -14063,6 +14063,10 @@ \section{Normalization}
 Schemes, Lemmas \ref{lemma-compose-after-separated},
 \ref{schemes-lemma-quasi-separated-coproduct}, and
 \ref{schemes-lemma-affine-separated}).
+This morphism induces a bijection on irreducible components
+almost by definition
+(Topology, Lemma
+\ref{topology-lemma-characterize-bijection-irreducible-components-sober}).
 
 \begin{definition}
 \label{definition-normalization}
@@ -14231,7 +14235,9 @@ \section{Normalization}
 The morphism $\nu$ is integral by Lemma \ref{lemma-characterize-normalization}.
 By Lemma \ref{lemma-normal-normalization} the
 morphism $Y \to X^\nu$ induces a bijection on irreducible components,
-and by construction of $Y$ this implies that $X^\nu \to X$ induces
+and by Topology, Lemma
+\ref{topology-lemma-bijection-irreducible-components-2-out-of-3},
+this implies that $X^\nu \to X$ induces
 a bijection on irreducible components. By construction $f : Y \to X$
 is dominant, hence also $\nu$ is dominant. Since an integral morphism is
 closed (Lemma \ref{lemma-integral-universally-closed}) this implies that
diff --git a/topology.tex b/topology.tex
index d949d14eb..11919a9f0 100644
--- a/topology.tex
+++ b/topology.tex
@@ -1037,6 +1037,30 @@ \section{Irreducible components}
 $Z\mapsto \overline{f(Z)}$ is a bijection between sets of irreducible components.
 \end{definition}
 
+\begin{lemma}
+\label{lemma-bijection-irreducible-components-2-out-of-3}
+Let $f:X\to Y$, $g:Y\to Z$ be continuous maps of topological spaces.
+\begin{enumerate}
+	\item If $f,g$ induce a bijection on irreducible components,
+	then so does $g\circ f$.
+	\item If $f,g\circ f$ induce a bijection on irreducible components,
+	then so does $g$.
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+First note that the composition of the map
+$F\subset X\mapsto \overline{f(F)}\subset Y$ with the map
+$T\subset Y\mapsto \overline{g(T)}$ equals the map
+$F\subset X\mapsto\overline{g(f(F))}$ (use continuity of $g$).
+Thus, the first point is clear. To prove the second point,
+it suffices to show that if $T\subset Y$ is an irreducible component,
+then $\overline{g(T)}$ is an irreducible component.
+There is an irreducible component $T\subset X$ such that $\overline{f(F)}=T$.
+Hence, $\overline{g(T)}=\overline{g(\overline{f(T)})}=\overline{g(f(F))}$
+is an irreducible component.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-bijection-irreducible-components-local-target}
 Let $f:X\to Y$ be a continuous map of topological spaces.