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/categories.tex b/categories.tex index 6cacb0be8..a43761537 100644 --- a/categories.tex +++ b/categories.tex @@ -755,6 +755,130 @@ \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} +\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 +$$ +\xymatrix{ +X' \ar[r]^(.37){\Delta_{X'/S'}}\ar[d] & X'\times_{S'}X'\ar[d]\\ +X \ar[r]^(.35){\Delta_{X/S}} & X\times_{S}X +} +$$ +\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/morphisms.tex b/morphisms.tex index 76d88114c..20ff244e4 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. @@ -10663,8 +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} @@ -10684,8 +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} @@ -10717,7 +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} @@ -13688,12 +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} @@ -13776,7 +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 @@ -13811,14 +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} @@ -13977,7 +14060,13 @@ \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}). +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} @@ -14057,13 +14146,17 @@ \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} @@ -14142,7 +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/properties.tex b/properties.tex index d38ed6bef..d6f24c407 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-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 + 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. diff --git a/schemes.tex b/schemes.tex index 759943ccc..ac5f0807c 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} @@ -3326,6 +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 @@ -3438,36 +3477,6 @@ \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 +3494,43 @@ \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{categories-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. @@ -3565,19 +3610,44 @@ \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 @@ -3585,20 +3655,8 @@ \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} @@ -4072,18 +4130,104 @@ \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} +(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{lemma-closed-local-target}. + +\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} + +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 $$ \xymatrix{ @@ -4092,11 +4236,52 @@ \section{Separation axioms} Y \ar[r]^-{\Delta_{Y/S}} & Y \times_S Y } $$ -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. @@ -4221,15 +4406,9 @@ \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{categories-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 +4464,11 @@ \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{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). \end{proof} \begin{lemma} @@ -4375,6 +4537,74 @@ \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..11919a9f0 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 @@ -979,6 +1028,111 @@ \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-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. +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. @@ -1060,6 +1214,38 @@ \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 @@ -5209,7 +5395,33 @@ \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}