HW5

If a topological space can be covered by finitely many connected open sets, and every pair of these sets has a nonempty intersection, then the space is connected.

\begin{proof} Assume that $X$ is a topological space can be covered by finitely many connected open sets, and assume that $X$ can be written as $X=U_1\bigsqcup U_2$ where $U_1$ and $U_2$ are open. Assume that $X={\cup }_{i=1}^n{X}_i$ and $X_i\cap X_j\ne \varnothing$ for any $i,j$. Since $U_i$ are open sets, $U_1\cap X_i$ and $U_2\cap X_i$ are open sets of $X_i$ and $X_i=(U_1\cap X_i)\bigsqcup (U_2\cap X_i)$. Since $X_i$ are connected, either $X_i\subseteq U_1$ or $X_i\subseteq U_2$. If $U_1\supseteq {\cup }_{i\in I}{X}_i$ and $U_2\supseteq {\cup }_{j\in J}{X}_j$, then $I\cap J=\varnothing$ because every pair of $X_i$‘s has a nonempty intersection. Thus there exists at least one of $U_1,U_2$ does not contain any $X_i$, which deduces that either $U_1$ or $U_2$ equals $\varnothing$. It deduces that the space is connected. \end{proof}

This is a statement we used in ^c0400a.
Let $Y$ be a topological space, $Z\subseteq Y$ be a subset, and $V\subseteq Y$ be an open subset. If a point $z_0\in V$ also belongs to $\overline{Z}$, then $z_0$ belongs to the closure of the intersection $Z\cap V$ taken within the subspace topology of $V$ (denoted ${\overline{Z\cap V}}^V$).

\begin{proof} It suffices to show that every open neighborhood $N\subseteq V$ of $z_0$ in $V$ has a non-empty intersection with the set $Z\cap V$.

Let $N$ be an open neighborhood of $z_0$ in the subspace topology of $V$, then $N$ is also open in $Y$. Since $z_0\in \overline{Z}$, one have $N\cap Z\ne \varnothing$. Notice that $N\cap Z=N\cap (Z\cap V)$, so we finish the proof. \end{proof}

Let $X$ be a variety, and let $U=\mathrm{Spec}(R)$, $V=\mathrm{Spec}(S)$ be two affine open subvarieties of $X$. In ^jye8sb, we have shown that $U\cap V$ is isomorphic to the affine variety $Z^{\prime }=\Delta (X)\cap (U\times V)$ via the mutually inverse morphisms:

• $\tau :U\cap V\to Z^{\prime }$, defined by $\tau (x)=(x,x)$
• $\theta :Z^{\prime }\to U\cap V$, defined by $\theta (x,x)=x$ (restriction of the first projection $p_1:U\times V\to U$).

Define $I(Z^{\prime })$ as the corresponding ideal of $Z^{\prime }$ in $R{\otimes }_{k}S$, where $R{\otimes }_{k}S$ is generated by $\{f\otimes g:f\in R,g\in S\}$. Then $\Gamma (Z^{\prime })\simeq R\otimes S/I(Z^{\prime })$. The task is to explicitly describe the ring isomorphism induced by $\tau$ and $\theta$.

\begin{proof} For $\overline{f\otimes g}\in \Gamma (Z^{\prime })$, the pullback is ${\tau }^{\ast }(\overline{f\otimes g})=\overline{f\otimes g}\circ \tau$. Evaluating at $w\in U\cap V$, there is

$${\tau }^{\ast }(\overline{f\otimes g})(w)=(\overline{f\otimes g})(\tau (w))=(\overline{f\otimes g})(w,w)=f(w)g(w).$$

Action of ${\theta }^{\ast }$: For $k\in \Gamma (U\cap V)$, the pullback is ${\theta }^{\ast }(k)=k\circ \theta$. Evaluating at $(w,w)\in Z^{\prime }$ with $w\in U\cap V$, there is

$${\theta }^{\ast }(k)(w,w)=k(\theta (w,w))=k(w).$$

It is easy to check ${\tau }^{\ast }$ and ${\theta }^{\ast }$ are inverse. \end{proof}

Let $X$ be a variety (or prevariety) over a field $k$. Let $f:X\to X$ be a morphism such that the map on the underlying topological space is the identity map, i.e., $f(x)=x$ for all points $x\in X$. Prove that $f$ is the identity morphism ${\mathrm{id}}_X$. This statement is used in ^jye8sb.

\begin{proof} Let $U\subseteq X$ be any open set, and let $s\in \Gamma (U,{\underline{o}}_X)$. Since $f(x)=x$, the preimage $f^{-1}(U)$ is simply $U$ and $f^{\ast }(s)=s\circ f=s\in \Gamma (U,{\underline{o}}_X)$. Therefore, the map $f^{\ast }:\Gamma (U,{\underline{o}}_X)\to \Gamma (U,{\underline{o}}_X)$ is the identity homomorphism. Since the map on points $f$ is the identity map and $f^{\ast }$ is the identity pullback map, we know $f={\mathrm{id}}_X$. \end{proof}

If $Z$ is an irreducible topological space and $U$ is an open subset of $X$, then $Z\cap U$ is irreducible.

\begin{proof} To show $Y:=Z\cap U$ is irreducible, it suffices to show for any two nonempty open sets $V_1,V_2\subseteq Y$, there is $V_1\cap V_2\ne \varnothing$. Since $X$ is a topological space and $Y$ is a topological subspace, there exist open subsets $O_1,O_2$ of $X$ such that $V_1=O_1\cap Y$, $V_2=O_2\cap Y$.

Define $O_1^{\prime }=O_1\cap Z$ and $O_2^{\prime }=O_2\cap Z$, then $O_1^{\prime },O_2^{\prime }$ are open sets in $Z$. Since $O_1^{\prime }\cap U=O_1\cap Z\cap U=V_1$ and $O_2^{\prime }\cap U=V_2$, they are nonempty open subsets of $Z$. Since $Z$ is irreducible, we have $(O_1^{\prime }\cap U)\cap (O_2^{\prime }\cap U)\ne \varnothing$, that is, $V_1\cap V_2\ne \varnothing$. Therefore, $Z\cap U$ is irreducible. \end{proof}

For a morphism $f:X\to Y$ and $Z=\overline{f(X)}$, the restriction $f^{\prime }:X\to Z$ is also a morphism.

\begin{proof} For any open subset $O\subseteq Z$, there exists open subset $U$ of $Y$ such that $O=Z\cap U$. Since $f$ is continuous, $f^{-1}(U)$ is open. Since $Z=\overline{f(X)}$, there is $f^{-1}(U\setminus Z)=\varnothing$ and so $f^{-1}(O)=f^{-1}(U)$ is open. Thus $f^{\prime }$ is continuous.

It remains to show for any open subset $V\subseteq Z$ and $g\in \Gamma (V,{\underline{o}}_Z)$, one have $g\circ f^{\prime }\in \Gamma (f^{\prime -1}(V),{\underline{o}}_X)$. Note that there exists an open subset $W\subseteq Y$ such that $W\cap Z=V$ and $f^{\prime -1}(V)=f^{-1}(W)$. For any $x\in f^{-1}(W)$, define $y=f^{\prime }(x)=f(x)\in V$. Since $g\in \Gamma (V,{\underline{o}}_Z)$, there exists an open neighborhood $W_y\subseteq Y$ and $h\in \Gamma (W_y,{\underline{o}}_Y)$ such that $g{|}_{V\cap W_y}=h{|}_{V\cap W_y}$. Since $f$ is a morphism, $h\circ f\in \Gamma (f^{-1}(W_y),{\underline{o}}_X)$. Take $N_x=f^{-1}(W_y\cap W)$, which is an open neighborhood of $x$, and then $g\circ f^{\prime }=g\circ f=h\circ f=h\circ f^{\prime }$ over $N_x$ and so $g\circ f^{\prime }\in \Gamma (N_x,{\underline{o}}_X)$. By the arbitrary of $x$, $g\circ f^{\prime }\in \Gamma (f^{\prime -1}(V),{\underline{o}}_X)$. Now we finish the proof. \end{proof}

A compact complex manifold is universal closed.

\begin{proof} It suffices to show for any variety $Y$, $p_r:X\times Y\to Y$ is a closed map.

Assume that $U\subseteq X\times Y$ is a closed subset of $X\times Y$, we aim to show $p_r(U)$ is a closed set. For any limit point $y\in Y$ of $p_r(U)$, there exists a sequence $y_n\to y$. Then there exists $\{x_n\}$ such that $(x_n,y_n)\in U$. Since $X$ is compact, $(x_n)$ has a subsequence $(x_{n_k})$ such that $x_{n_k}\to x$ and so $(x_{n_k},y_{n_k})\to (x,y)$. Since $U$ is closed, $(x,y)\in U$ and so $y\in p_r(U)$. Therefore, $p_r(Y)$ is closed. \end{proof}