A \vocab{presheaf}$\mathcal{G}$ of sets (or rings, (abelian) groups) on $X$ associates a set (or rings, or (abelian) group) $\mathcal{G}(U)$ to every open subset $U$ of $X$, and a map (or ring or group homomorphism) $\mathcal{G}(U)\xrightarrow{r_{U,V}}\mathcal{G}(V)$ to every inclusion $V \subseteq U$ of open subsets of $X$ such that $r_{U,W}= r_{V,W} r_{U,V}$ for inclusions $U \subseteq V \subseteq W$ of open subsets.
Elements of $\mathcal{G}(U)$ are often called \vocab{sections} of $\mathcal{G}$ on $U$ or \vocab{global sections} when $U = X$.
Let $U \subseteq X$ be open and $U =\bigcup_{i \in I} U_i$ an open covering.
A family $(f_i)_{i \in I}\in\prod_{i \in I}\mathcal{G}(U_i)$ is called \vocab[Sections!compatible]{compatible} if $r_{U_i, U_i \cap U_j}(f_i)= r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$.
A presheaf is called \vocab[Presheaf!separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is injective for all such $U$ and $(U_i)_{i \in I}$.\footnote{This also called ``locality''.}
It satisfies \vocab{gluing} if $\phi_{U, (U_i)_{i \in I}}$ is surjective.
A presheaf is called a \vocab{sheaf} if it is separated and satisfies gluing.
The bijectivity of the $\phi_{U, (U_i)_{i \in I}}$ is called the \vocab{sheaf axiom}.
\end{definition}
\begin{trivial}+
A presheaf is a contravariant functor $\mathcal{G} : \mathcal{O}(X)\to C$ where $\mathcal{O}(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups.
\end{trivial}
\begin{definition}
A subsheaf $\mathcal{G}'$ is defined by subsets (resp. subrings or subgroups) $\mathcal{G}'(U)\subseteq\mathcal{G}(U)$ for all open $U \subseteq X$ such that the sheaf axioms still hold.
\end{definition}
\begin{remark}
If $\mathcal{G}$ is a sheaf on $X$ and $\Omega\subseteq X$ open, then $\mathcal{G}\defon{\Omega}(U)\coloneqq\mathcal{G}(U)$ for open $U \subseteq\Omega$ and $r_{U,V}^{(\mathcal{G}\defon{\Omega})}(f)\coloneqq r_{U,V}^{(\mathcal{G})}(f)$ is a sheaf of the same kind as $\mathcal{G}$ on $\Omega$.
\end{remark}
\begin{remark}
The notion of restriction of a sheaf to a closed subset, or of preimages under general continuous maps, can be defined but this is a bit harder.
\end{remark}
\begin{notation}
It is often convenient to write $f \defon{V}$ instead of $r_{U,V}(f)$.
\end{notation}
\begin{remark}
Applying the \vocab{sheaf axiom} to the empty covering of $U =\emptyset$, one finds that $\mathcal{G}(\emptyset)=\{0\}$.
\end{remark}
\subsubsection{Examples of sheaves}
\begin{example}
Let $G$ be a set and let $\mathfrak{G}(U)$ be the set of arbitrary maps $U \xrightarrow{f} G$. We put $r_{U,V}(f)= f\defon{V}$.
It is easy to see that this defines a sheaf.
If $\cdot$ is a group operation on $G$, then $(f\cdot g)(x)\coloneqq f(x)\cdot g(x)$ defines the structure of a sheaf of group on $\mathfrak{G}$.
Similarly, a ring structure on $G$ can be used to define the structure of a sheaf of rings on $\mathfrak{G}$.
\end{example}
\begin{example}
If in the previous example $G$ carries a topology and $\mathcal{G}(U)\subseteq\mathfrak{G}(U)$ is the subset (subring, subgroup) of continuous functions $U \xrightarrow{f} G$, then $\mathcal{G}$ is a subsheaf of $\mathfrak{G}$, called the sheaf of continuous $G$-valued functions on (open subsets of) $X$.
\end{example}
\begin{example}
If $X =\R^n$, $\mathbb{K}\in\{\R, \C\}$ and $\mathcal{O}(U)$ is the sheaf of $\mathbb{K}$-valued $C^{\infty}$-functions on $U$, then $\mathcal{O}$ is a subsheaf of the sheaf (of rings) of $\mathbb{K}$-valued continuous functions on $X$.
\end{example}
\begin{example}
If $X =\C^n$ and $\mathcal{O}(U)$ the set of holomorphic functions on $X$, then $\mathcal{O}$ is a subsheaf of the sheaf of $\C$-valued $C^{\infty}$-functions on $X$.
\end{example}
\subsubsection{The structure sheaf on a closed subset of $\mathfrak{k}^n$}
Let $X \subseteq\mathfrak{k}^n$ be open. Let $R =\mathfrak{k}[X_1,\ldots,X_n]$.
\begin{definition}\label{structuresheafkn}
For open subsets $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi}\mathfrak{k}$ such that every $x \in U$ has a neighbourhood $V$ such that there are $f,g \in R$ such that for $y \in V$ we have $g(y)\neq0$ and $\phi(y)=\frac{f(y)}{g(y)}$.
\end{definition}
\begin{remark}\label{structuresheafcontinuous}
$\mathcal{O}_X$ is a subsheaf (of rings) of the sheaf of $\mathfrak{k}$-valued functions on $X$.
The elements of $\mathcal{O}_X(U)$ are continuous:
Let $M \subseteq\mathfrak{k}$ be closed. We must show the closedness of $N \coloneqq\phi^{-1}(M)$ in $U$. For $M =\mathfrak{k}$ this is trivial. Otherwise $M$ is finite and we may assume $M =\{t\}$ for some $t \in\mathfrak{k}$. For $x \in U$, there are open $V_x \subseteq U$ and $f_x, g_x \in R$ such that $\phi=\frac{f_x}{g_x}$ on $V_x$.
Then $N \cap V_x = V(f_x - t\cdot g_x)\cap V_x)$ is closed in $V_x$. As the $V_x$ cover $U$ and $U$ is quasi-compact, $N$ is closed in $U$.
\end{remark}
\begin{proposition}\label{structuresheafri}
Let $X = V(I)$ where $I =\sqrt{I}\subseteq R$ is an ideal. Let $A = R / I$. Then
\begin{align}
\phi: A &\longrightarrow\mathcal{O}_X(X) \\
f \mod I &\longmapsto f\defon{X}
\end{align}
is an isomorphism.
\end{proposition}
\begin{proof}
It is easy to see that the map $A \to\mathcal{O}_X(X)$ is well-defined and a ring homomorphism.
Its injectivity follows from the Nullstellensatz and $I =\sqrt{I}$ (\ref{hns3}).
Let $\phi\in\mathcal{O}_X(X)$. for $x \in X$, there are an open subset $U_x \subseteq X$ and $f_x, g_x \in R$ such that $\phi=\frac{f_x}{g_x}$ on $U_x$.
\subsubsection{The structure sheaf on closed subsets of $\mathbb{P}^n$}
Let $X \subseteq\mathbb{P}^n$ be closed and $R_\bullet=\mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading.
\begin{definition}\label{structuresheafpn}
For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi}\mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap\Vp(g)=\emptyset$ and $\phi(y)=\frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y =[y_0,\ldots,y_n]\in W$.
\end{definition}
\begin{remark}
This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$.
Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ with $\mathbb{P}^n \setminus\Vp(X_0)$, one has $\mathcal{O}_X \defon{X \setminus\Vp(X_0)}=\mathcal{O}_{X \cap\mathbb{A}^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n])=\frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n]\in W$.
Conversely if $\phi([1,y_1,\ldots,y_n])=\frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap\mathbb{A}^n$ then
$\phi([y_0,\ldots,y_n])=\frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n)\coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n)= X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in\N$.
\end{remark}
\begin{remark}
It follows from the previous remark and the similar result in the affine case that the elements of $\mathcal{O}_X(U)$ are continuous on $U \setminus V(X_0)$.
Since the situation is symmetric in the homogeneous coordinates, they are continuous on all of $U$.
\end{remark}
The following is somewhat harder than in the affine case:
\begin{proposition}
If $X$ is connected (e.g. irreducible), then the elements of $\mathcal{O}_X\left( X \right)$ are constant functions on $X$.
\end{proposition}
% Lecture 14
\subsection{The notion of a category}
\begin{definition}
A \vocab{category}$\mathcal{A}$ consists of:
\begin{itemize}
\item A class $\Ob\mathcal{A}$ of \vocab[Objects]{objects of $\mathcal{A}$}.
\item For two arbitrary objects $A, B \in\Ob\mathcal{A}$, a \textbf{set}$\Hom_\mathcal{A}(A,B)$ of \vocab[Morphism]{morphisms for $A$ to $B$ in $\mathcal{A}$}.
\item A map $\Hom_\mathcal{A}(B,C)\times\Hom_\mathcal{A}(A,B)\xrightarrow{\circ}\Hom_\mathcal{A}(A,C)$, the composition of morphisms, for arbitrary triples $(A,B,C)$ of objects of $\mathcal{A}$.
\end{itemize}
The following conditions must be satisfied:
\begin{enumerate}[A]
\item For morphisms $A \xrightarrow{f} B\xrightarrow{g} C \xrightarrow{h} D$, we have $h \circ(g \circ f)=(h \circ g)\circ f$.
\item For every $A \in\Ob(\mathcal{A})$, there is an $\Id_A \in\Hom_{\mathcal{A}}(A,A)$ such that $\Id_A \circ f = f$ (reps. $g \circ\Id_A = g$) for arbitrary morphisms $B \xrightarrow{f} A$ (reps. $A \xrightarrow{g} C).$
\end{enumerate}
A morphism $X \xrightarrow{f} Y$ is called an \vocab[Isomorphism]{isomorphism (in $\mathcal{A}$)} if there is a morphism $Y \xrightarrow{g} X$ (called the \vocab[Inverse morphism]{inverse $f^{-1}$ of $f$)} such that $g \circ f =\Id_X$ and $f \circ g =\Id_Y$.
\end{definition}
\begin{remark}
\begin{itemize}
\item The distinction between classes and sets is important here.
\item We will usually omit the composition sign $\circ$.
\item It is easy to see that $\Id_A$ is uniquely determined by the above condition $B$, and that the inverse $f^{-1}$ of an isomorphism $f$ is uniquely determined.
\end{itemize}
\end{remark}
\subsubsection{Examples of categories}
\begin{example}
\begin{itemize}
\item The category of sets.
\item The category of groups.
\item The category of rings.
\item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras
\item The category of topological spaces
\item The category $\Var_\mathfrak{k}$ of varieties over $\mathfrak{k}$ (see \ref{defvariety})
\item If $\mathcal{A}$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op)=\Ob(\mathcal{A})$ and $\Hom_{\mathcal{A}\op}(X,Y)=\Hom_\mathcal{A}(Y,X)$.
\end{itemize}
In most of these cases, isomorphisms in the category were just called `isomorphism'. The isomorphisms in the category of topological spaces are the homeomophisms.
\end{example}
\subsubsection{Subcategories}
\begin{definition}[Subcategories]
A \vocab{subcategory} of $\mathcal{A}$ is a category $\mathcal{B}$ such that $\Ob(\mathcal{B})\subseteq\Ob(\mathcal{A})$, such that $\Hom_\mathcal{B}(X,Y)\subseteq\Hom_\mathcal{A}(X,Y)$ for objects $X$ and $Y$ of $\mathcal{B}$, such that for every object $X \in\Ob(\mathcal{B})$, the identity $\Id_X$ of $X$ is the same in $\mathcal{B}$ as in $\mathcal{A}$, and such that for composable morphisms in $\mathcal{B}$, their compositions in $\mathcal{A}$ and $\mathcal{B}$ coincide.
We call $\mathcal{B}$ a \vocab{full subcategory} of $\mathcal{A}$ if in addition $\Hom_\mathcal{B}(X,Y)=\Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in\Ob(\mathcal{B})$.
\end{definition}
\begin{example}
\begin{itemize}
\item The category of abelian groups is a full subcategory of the category of groups.
It can be identified with the category of $\Z$-modules.
\item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules.
\item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$.
\item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$.
\end{itemize}
\end{example}
\subsubsection{Functors and equivalences of categories}
\begin{definition}
A \vocab[Functor!covariant]{(covariant) functor} (resp. \vocab[Functor!contravariant]{contravariant functor}) between categories $\mathcal{A}\xrightarrow{F}\mathcal{B}$ is a map $\Ob(\mathcal{A})\xrightarrow{F}\Ob(\mathcal{B})$ with a family of maps $\Hom_\mathcal{A}(X,Y)\xrightarrow{F}\Hom_\mathcal{B}(F(X),F(Y))$ (resp. $\Hom_\mathcal{A}(X,Y)\xrightarrow{F}\Hom_\mathcal{B}(F(Y),F(X))$ in the case of contravariant functors), where $X$ and $Y$ are arbitrary objects of $\mathcal{A}$, such that the following conditions hold:
\begin{itemize}
\item$F(\Id_X)=\Id_{F(X)}$
\item For morphisms $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathcal{A}$, we have $F(gf)= F(g)F(f)$ ( resp. $F(gf)= F(f)F(g)$)
\end{itemize}
A functor is called \vocab[Functor!essentially surjective]{essentially surjective} if every object of $\mathcal{B}$ is isomorphic to an element of the image of $\Ob(\mathcal{A})\xrightarrow{F}\Ob(\mathcal{B})$.
A functor is called \vocab[Functor!full]{full} (resp. \vocab[Functor!faithful]{faithful}) if it induces surjective (resp. injective) maps between sets of morphisms.
It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective.
\end{definition}
\begin{example}
\begin{itemize}
\item There are \vocab[Functor!forgetful]{forgetful functors} from rings to abelian groups or from abelian groups to sets which drop the multiplicative structure of a ring or the group structure of a group.
\item If $\mathfrak{k}$ is any vector space there is a contravariant functor from $\mathfrak{k}$-vector spaces to itself sending $V$ to its dual vector space $V\subseteq$ and $V \xrightarrow{f} W$ to the dual linear map $W^{\ast}\xrightarrow{f^{\ast}} V^{\ast}$.
When restricted to the full subcategory of finite-dimensional vector spaces it becomes a contravariant self-equivalence of that category.
\item The embedding of a subcategory is a faithful functor. In the case of a full subcategory it is also full.
An \vocab{algebraic variety} or \vocab{prevariety} over $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ such that for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$\footnote{By the result of \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without altering the definition.} and a homeomorphism $V_x \xrightarrow{\iota_x} U_x$ such that for every open subset $V \subseteq U_x$ and every function $V\xrightarrow{f}\mathfrak{k}$, we have $f \in\mathcal{O}_X(V)\iff\iota^{\ast}_x(f)\in\mathcal{O}_{Y_x}(\iota_x^{-1}(V))$,
In this, the \vocab{pull-back}$\iota_x^{\ast}(f)$ of $f$ is defined by $(\iota_x^{\ast}(f))(\xi)\coloneqq f(\iota_x(\xi))$.
A morphism $(X, \mathcal{O}_X)\to(Y, \mathcal{O}_Y)$ of varieties is a continuous map $X \xrightarrow{\phi} Y$ such that for all open $U \subseteq Y$ and $f \in\mathcal{O}_Y(U)$, $\phi^{\ast}(f)\in\mathcal{O}_X(\phi^{-1}(U))$.
An isomorphism is a morphism such that $\phi$ is bijective and $\phi^{-1}$ also is a morphism of varieties.
\end{definition}
\begin{example}
\begin{itemize}
\item If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then $(U, \mathcal{O}_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties.
\item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
\item If $X = V(X^2- Y^3)\subseteq\mathfrak{k}^2$ then $\mathfrak{k}\xrightarrow{t \mapsto(t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties.
% TODO
\item The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of varieties.
\item$X\xrightarrow{\Id_X} X$ is a morphism of varieties.
\end{itemize}
\end{example}
\subsubsection{The category of affine varieties}
\begin{lemma}\label{localinverse}
Let $X$ be any $\mathfrak{k}$-variety and $U \subseteq X$ open.
\begin{enumerate}[i)]
\item All elements of $\mathcal{O}_X(U)$ are continuous.
\item If $U \subseteq X$ is open, $U \xrightarrow{\lambda}\mathfrak{k}$ any function and every $x \in U$ has a neighbourhood $V_x \subseteq U$ such that $\lambda\defon{V_x}\in\mathcal{O}_X(V_x)$, then $\lambda\in\mathcal{O}_X(U)$.
\item If $\vartheta\in\mathcal{O}_X(U)$ and $\vartheta(x)\neq0$ for all $x \in U$, then $\vartheta\in\mathcal{O}_X(U)^{\times}$.
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}[i)]
\item The property is local on $U$, hence it is sufficient to show it in the quasi-affine case. This was done in \ref{structuresheafcontinuous}.
\item For the second part, let $\lambda_x \coloneqq\lambda\defon{V_x}$.
We have $\lambda_x\defon{V_x \cap V_y}=\lambda\defon{V_x \cap V_y}=\lambda_y \defon{V_x \cap V_y}$.
The $V_x$ cover $U$. By the sheaf axiom for $\mathcal{O}_X$ there is $\ell\in\mathcal{O}_X(U)$ with $\ell\defon{V_x}=\lambda_x$. It follows that $\ell=\lambda$.
\item By the definition of variety, every $x \in U$ has a quasi-affine neighbourhood $V \subseteq U$. We can assume $U$ to be quasi-affine and $X = V(I)\subseteq\mathfrak{k}^n$, as the general assertion follows by an application of ii).
If $x \in U$ there are a neighbourhood $x \in W \subseteq U$ and $a,b \in R =\mathfrak{k}[X_1,\ldots,X_n]$ such that $\vartheta(y)=\frac{a(y)}{b(y)}$ for $y \in W$, with $b(y)\neq0$.
Then $a(x)\neq0$ as $\vartheta(x)\neq0$. Replacing $W$ by $W \setminus V(a)$, we may assume that $a$ has no zeroes on $W$.
Then $\lambda(y)=\frac{b(y)}{a(y)}$ for $y \in W$ has a non-vanishing denominator and $\lambda\in\mathcal{O}_X(U)$.
We have $\lambda\cdot\vartheta=1$, thus $\vartheta\in\mathcal{O}_X(U)^{\times}$.
\end{enumerate}
\end{proof}
\begin{proposition}[About affine varieties]
\label{propaffvar}
\begin{itemize}
\item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map
restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of $\Alg_\mathfrak{k}$,
having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A =\{0\}$ as objects.
\end{itemize}
\end{proposition}
\begin{remark}
It is clear that $\nil(\mathcal{O}_X(X))=\{0\}$ for arbitrary varieties. For general varieties it is however not true that $\mathcal{O}_X(X)$ is a $\mathfrak{k}$-algebra of finite type.
There are counterexamples even for quasi-affine $X$. %TODO
If, however, $X$ is affine, we may assume w.l.o.g. that $X = V(I)$ where $I =\sqrt{I}\subseteq R$ is an ideal with $R =\mathfrak{k}[X_1,\ldots,X_n]$.
Then $\mathcal{O}_X(X)\cong R / I$ (see \ref{structuresheafri}) is a $\mathfrak{k}$-algebra of finite type.
\end{remark}
\begin{proof}
It suffices to investigate $\phi$ when $Y$ is an open subset of $V(I)\subseteq\mathfrak{k}^n$, where $I =\sqrt{I}\subseteq R$ is an ideal and $Y = V(I)$ when $Y$ is affine.
Let $(f_1,\ldots,f_n)$ be the components of $X \xrightarrow{f} Y \subseteq\mathfrak{k}^n$. Let $Y \xrightarrow{\xi_i}\mathfrak{k}$ be the $i$-th coordinate.
By definition $f_i = f^{\ast}(\xi_i)$. Thus $f$ is uniquely determined by $\mathcal{O}_Y(Y)\xrightarrow{f^{\ast}}\mathcal{O}_X(X)$.
Conversely, let $Y = V(I)$ and $\mathcal{O}_Y(Y)\xrightarrow{\phi}\mathcal{O}_X(X)$ be a morphism of $\mathfrak{k}$-algebras. Define $f_i \coloneqq\phi(\xi_i)$ and consider $X \xrightarrow{f =(f_1,\ldots,f_n)} Y\subseteq\mathfrak{k}^n$.
\begin{claim}
$f$ has image contained in $Y$.
\end{claim}
\begin{subproof}
For $x \in X, \lambda\in I$ we have $\lambda(f(x))=(\phi(\lambda\mod I))(x)=0$ as $\phi$ is a morphism of $\mathfrak{k}$-algebras.
Thus $f(x)\in V(I)= Y$.
\end{subproof}
\begin{claim}
$f$ is a morphism in $\Var_\mathfrak{k}$
\end{claim}
\begin{subproof}
For open $\Omega\subseteq Y, U = f^{-1}(\Omega)=\{x \in X | \forall\lambda\in J ~ (\phi(\lambda))(x)\neq0\}$ is open in $X$, where $Y \setminus\Omega= V(J)$.
If $\lambda\in\mathcal{O}_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v)=\frac{a(v)}{b(v)}$ and $b(v)\neq0$ for all $v \in V$.
Let $W \coloneqq f^{-1}(V)$. Then $\alpha\coloneqq\phi(a)\defon{W}\in\mathcal{O}_X(W)$, $\beta\coloneqq\phi(b)\defon{W}\in\mathcal{O}_X(W)$.
By the second part of \ref{localinverse}$\beta\in\mathcal{O}_X(W)^{\times}$ and $f^{\ast}(\lambda)\defon{W}=\frac{\alpha}{\beta}\in\mathcal{O}_X(W)$.
The first part of \ref{localinverse} shows that $f^{\ast}(\lambda)\in\mathcal{O}_X(U)$.
\end{subproof}
By definition of $f$, we have $f^{\ast}=\phi$. This finished the proof of the first point.
\begin{claim}
The functor in the second part maps affine varieties to objects of $\mathcal{A}$ and is essentially surjective.
\end{claim}
\begin{subproof}
It follows from the remark that the functor maps affine varieties to objects of $\mathcal{A}$.
If $A \in\Ob(\mathcal{A})$ then $ A /\mathfrak{k}$ is of finite type, thus $A \cong R / I$ for some $n$.
Since $\nil(A)=\{0\}$ we have $I =\sqrt{I}$, as for $x \in\sqrt{I}$, $x \mod I \in\nil(R / I)\cong\nil(A)=\{0\}$.
Thus $A \cong\mathcal{O}_X(X)$ where $X = V(I)$.
\end{subproof}
Fullness and faithfulness of the functor follow from the first point.
\end{proof}
\begin{remark}
Note that giving a contravariant functor $\mathcal{C}\to\mathcal{D}$ is equivalent to giving a functor $\mathcal{C}\to\mathcal{D}\op$. We have thus shown that the category of affine varieties is equivalent to $\mathcal{A}\op$, where $\mathcal{A}\subsetneq\Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A)=\{0\}$.
\end{remark}
\subsubsection{Affine open subsets are a topology base}
\begin{definition}
A set $\mathcal{B}$ of open subsets of a topological space $X$ is called a \vocab{topology base} for $X$ if every open subset of $X$ can be written as a (possibly empty) union of elements of $\mathcal{B}$.
\end{definition}
\begin{fact}
If $X$ is a set, then $\mathcal{B}\subseteq\mathcal{P}(X)$ is a base for some topology on $X$ iff $X =\bigcup_{U \in\mathcal{B}} U$ and for arbitrary $U, V \in\mathcal{B}, U \cap V$ is a union of elements of $\mathcal{B}$.
\end{fact}
\begin{definition}
Let $X$ be a variety.
An \vocab{affine open subset} of $X$ is a subset which is an affine variety.
\end{definition}
\begin{proposition}\label{oxulocaf}
Let $X$ be an affine variety over $\mathfrak{k}$, $\lambda\in\mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$.
Then $U$ is an affine variety and the morphism $\phi: \mathcal{O}_X(X)_\lambda\to\mathcal{O}_X(U)$ defined by the restriction $\mathcal{O}_X(X)\xrightarrow{\cdot |_U }\mathcal{O}_X(U)$ and the universal property of the localization is an isomorphism.
\end{proposition}
\begin{proof}
Let $X$ be an affine variety over $\mathfrak{k}, \lambda\in\mathcal{O}_X(X)$ and $U = X \setminus V(\lambda)$. The fact that $\lambda\defon{U}\in\mathcal{O}_x(U)^{\times}$ follows from \ref{localinverse}.
Thus the universal property of the localization $\mathcal{O}_X(X)_\lambda$ can be applied to $\mathcal{O}_X(X)\xrightarrow{\cdot |_U}\mathcal{O}_X(U)$.
\[
\begin{tikzcd}
\mathcal{O}_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto\frac{x}{1}}&\mathcal{O}_X(X)_\lambda\arrow[dotted, bend left]{dl}{\existsone\phi}\\
X \arrow[hookrightarrow]{r}{}& U \arrow[swap]{u}{\sigma}&\mathcal{O}_X(U)
\end{tikzcd}
\]
For the rest of the proof, we may assume $X = V(I)\subseteq\mathfrak{k}^n$ where $I =\sqrt{I}\subseteq R \coloneqq\mathfrak{k}[X_1,\ldots,X_n]$ is an ideal.
Then $A \coloneqq\mathcal{O}_X(X)\cong R / I$ and there is $\ell\in R$ such that $\ell\defon{X}=\lambda$.
Let $Y = V(J)\subseteq\mathfrak{k}^{n+1}$ where $J \subseteq\mathfrak{k}[Z,X_1,\ldots,X_n]$ is generated by the elements of $I$ and $1- Z\ell(X_1,\ldots,X_n)$.
Then $\mathcal{O}_Y(Y)\cong\mathfrak{k}[Z,X_1,\ldots,X_n]/ J \cong A[Z]/(1-\lambda Z)\cong A_\lambda$.
By the proposition about affine varieties (\ref{propaffvar}), the morphism $\mathfrak{s}: \mathcal{O}_Y(Y)\cong A_\lambda\to\mathcal{O}_X(U)$ corresponds to a morphism $U \xrightarrow{\sigma} Y$.
We have $\mathfrak{s}(Z \mod J)=\lambda^{-1}$ and $\mathfrak{s}(X_i \mod J)= X_i \mod I$.
Thus $\sigma(x)=(\lambda(x)^{-1}, x)$ for $x \in U$.
Moreover, the projection $Y \xrightarrow{\pi_0} X$ dropping the $Z$-coordinate has image contained in $U$, as for $(z,x)\in Y$ the equation
\[
1 = z\lambda(x)
\]
implies $\lambda(x)\neq0$. It thus defines a morphism $Y \xrightarrow{\pi} U$ and by the description of $\sigma$ it follows that $\sigma\pi=\Id_U$.
Similarly it follows that $\sigma\pi=\Id_Y$. Thus, $\sigma$ and $\pi$ are inverse to each other.
\end{proof}
\begin{corollary}\label{affopensubtopbase}
The affine open subsets of a variety $X$ are a topology base on $X$.
\end{corollary}
\begin{proof}
Let $X = V(I)\subseteq\mathfrak{k}^n$ with $I =\sqrt{I}$. If $U \subseteq X$ is open then $X \setminus U = V(J)$ with $J \supseteq I$ and $U =\bigcup_{f \in J}(X \setminus V(f))$.
Thus $U$ is a union of affine open subsets. The same then holds for arbitrary quasi-affine varieties.
Let $X$ be any variety, $U \subseteq X$ open and $x \in U$.
By the definition of variety, $x$ has a neighbourhood $V_x$ which is quasi-affine, and replacing $V_x$ by $U \cap V_x$ which is also quasi-affine we may assume $V_x \subseteq U$.
$V_x$ is a union of its affine open subsets. Because $U$ is the union of the $V_x$, $U$ as well is a union of affine open subsets.
\end{proof}
% Lecture 14A TODO?
% Lecture 15
% CRTPROG
\subsection{Stalks of sheaves}
\begin{definition}[Stalk]
Let $\mathcal{G}$ be a presheaf of sets on the topological space $X$, and let $x \in X$.
The \vocab{stalk} (\vocab[Stalk]{Halm}) of $\mathcal{G}$ at $x$ is the set of equivalence classes of pairs $(U, \gamma)$, where $U$ is an open neighbourhood of $x$ and $\gamma\in\mathcal{G}(U)$
and the equivalence relation $\sim$ is defined as follows:
$( U , \gamma)\sim(V, \delta)$ iff there exists an open neighbourhood $W \subseteq U \cap V$ of $x$ such that $\gamma\defon{W}=\delta\defon{W}$.
If $\mathcal{G}$ is a presheaf of groups, one can define a groups structure on $\mathcal{G}_x$ by
Let $\gamma,\delta\in\mathcal{G}(U)$. If $\mathcal{G}$ is a sheaf\footnote{or, more generally, a separated presheaf} and if for all $x \in U$, we have $\gamma_x =\delta_x$, then $\gamma=\delta$.
In the case of a sheaf, the image of the injective map $\mathcal{G}(U)\xrightarrow{\gamma\mapsto(\gamma_x)_{x \in U}}\prod_{x \in U}\mathcal{G}_x$
is the set of all $(g_x)_{x \in U}\in\prod_{x \in U}\mathcal{G}_x $ satisfying the following \vocab{coherence condition}:
For every $x \in U$, there are an open neighbourhood $W_x \subseteq U$ of $x$ and $g^{(x)}\in\mathcal{G}(W_x)$ with $g_y^{(x)}= g_y$ for all $y \in W_x$.
\end{fact}
\begin{proof}
Because of $\gamma_x =\delta_x$, there is $x \in W_x \subseteq U$ open such that $\gamma\defon{W_x}=\delta\defon{W_x}$. As the $W_x$ cover $U$, $\gamma=\delta$ by the sheaf axiom.
\end{proof}
\begin{definition}
Let $\mathcal{G}$ be a sheaf of functions.
Then $\gamma_x$ is called the \vocab{germ} of the function $\gamma$ at $x$.
The \vocab[Germ!value at $x$]{value at $x$} of $g =(U, \gamma)/\sim\in\mathcal{G}_x$ defined as $g(x)\coloneqq\gamma(x)$, which is independent of the choice of the representative $\gamma$.
\end{definition}
\begin{remark}
If $\mathcal{G}$ is a sheaf of $C^{\infty}$-functions (resp. holomorphic functions), then $\mathcal{G}_x$ is called the ring of germs of $C^\infty$-functions (resp. of holomorphic functions) at $x$.
\end{remark}
\subsubsection{The local ring of an affine variety}
\begin{definition}
If $X$ is a variety, the stalk $\mathcal{O}_{X,x}$ of the structure sheaf at $x$ is called the \vocab{local ring} of $X$ at $x$.
This is indeed a local ring, with maximal ideal $\mathfrak{m}_x =\{f \in\mathcal{O}_{X,x} | f(x)=0\}$.
\end{definition}
\begin{proof}
By \ref{localring} it suffices to show that $\mathfrak{m}_x$ is a proper ideal, which is trivial, and that the elements of $\mathcal{O}_{X,x}\setminus\mathfrak{m}_x$ are units in $\mathcal{O}_{X,x}$.
Let $g =(U, \gamma)/\sim\in\mathcal{O}_{X,x}$ and $g(x)\neq0$.
$\gamma$ is Zariski continuous (first point of \ref{localinverse}). Thus $V(\gamma)$ is closed. By replacing $U$ by $U \setminus V(\gamma)$ we may assume that $\gamma$ vanishes nowhere on $U$.
By the third point of \ref{localinverse} we have $\gamma\in\mathcal{O}_X(U)^{\times}$.
$(\gamma^{-1})_x$ is an inverse to $g$.
\end{proof}
\begin{proposition}\label{proplocalring}
Let $X =\Va(I)\subseteq\mathfrak{k}^n$ be equipped with its usual structure sheaf, where $I =\sqrt{I}\subseteq R =\mathfrak{k}[X_1,\ldots,X_n]$ . Let $x \in X$ and $A =\mathcal{O}_X(X)\cong R / I$.
$\{P \in R | P(x)=0\}\text{\reflectbox{$\coloneqq$}}\fn_x \subseteq R$ is maximal, $I \subseteq\fn_x$ and $\mathfrak{m}_x \coloneqq\fn_x / I$ is the maximal ideal of elements of $A$ vanishing at $x$.
If $\lambda\in A \setminus\mathfrak{m}_x$, we have $\lambda_x \in\mathcal{O}_{X,x}^{\times}$, where $\lambda_x$ denotes the image under $A \cong\mathcal{O}_X(X)\to\mathcal{O}_{X,x}$.
By the universal property of the localization, there exists a unique ring homomorphism $A_{\mathfrak{m}_x}\xrightarrow{\iota}\mathcal{O}_{X,x}$
such that
\[
\begin{tikzcd}
A \arrow{r}{}\arrow{d}{\lambda\mapsto\lambda_x}& A_{\mathfrak{m}_x}\arrow[dotted, bend left]{ld}{\existsone\iota}\\
\mathcal{O}_{X,x}
\end{tikzcd}
\]
commutes.
The morphism $A_{\mathfrak{m}_x}\xrightarrow{\iota}\mathcal{O}_{X,x}$ is an isomorphism.
\end{proposition}
\begin{proof}
To show surjectivity, let $\ell=(U, \lambda)/\sim\in\mathcal{O}_{X,x}$, where $U$ is an open neighbourhood of $x$ in $X$.
We have $X \setminus U = V(J)$ where $J \subseteq A$ is an ideal. As $x \in U$ there is $f \in J$ with $f(x)\neq0$. Replacing $U $ by $X \setminus V(f)$ we may assume $U = X \setminus V(f)$.
By \ref{oxulocaf}, $\mathcal{O}_X(U)\cong A_f$, and $\lambda= f^{-n}\vartheta$ for some $n \in\N$ and $\vartheta\in A$.
Then $\ell=\iota(f^{-n}\vartheta)$ where the last fraction is taken in $A_{\mathfrak{m}_x}$.
Let $\lambda=\frac{\vartheta}{g}\in A_{\mathfrak{m}_x}$ with $\iota(\lambda)=0$.
It is easy to see that $\iota(\lambda)=(X \setminus V(g), \frac{\vartheta}{g})/\sim$.
Thus there is an open neighbourhood $U$ of $x$ in $X \setminus V(g)$ such that $\vartheta$ vanishes on $U$.
Similar as before there is $h \in A$ with $h(x)\neq0$ and $W = X \setminus V(h)\subseteq U$.
By the isomorphism $\mathcal{O}_X(W)\cong A_h$, there is $n \in\N$ with $h^{n}\vartheta=0$ in $A$. Since $h \not\in\mathfrak{m}_x$, $h$ is a unit and the image of $\vartheta$ in $A_{\mathfrak{m}_x}$ vanishes, implying $\lambda=0$.
\end{proof}
\subsubsection{Intersection multiplicities and Bezout's theorem}
\begin{definition}
Let $R =\mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in\mathbb{P}^{2}$.
Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G)\cap V(h)$.
Let $\ell\in R_1$ such that $\ell(x)\neq0$. Then $x \in U =\mathbb{P}^2\setminus V(\ell)$ and the rational functions $\gamma=\ell^{-g}G, \eta=\ell^{-h}H$ are elements of $\mathcal{O}_{\mathbb{P}^2}(U)$.
Let $I_x(G,H)\subseteq\mathcal{O}_{\mathbb{P}^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$.
\noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x}/ I_x(G,H))\text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$.
\end{definition}
\begin{remark}
If $\tilde\ell\in R_1$ also satisfies $\tilde\ell(x)\neq0$, then the image of $\tilde\ell/\ell$ under $\mathcal{O}_{\mathbb{P}^2}(U)\to\mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that the image of $\tilde\gamma=\tilde\ell^{-g} G$ in $\mathcal{O}_{\mathbb{P}^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$.
Thus $I_x(G,H)$ does not depend on the choice of $\ell\in R_1$ with $\ell(x)\neq0$.
\end{remark}
\begin{theorem}[Bezout's theorem]
In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G)\cap V(H)\subseteq\mathbb{P}^2$ has no irreducible component of dimension $\ge1$.
Then
\[
\sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh
\]
Thus, $V(G)\cap V(H)$ has $gh$ elements counted by multiplicity.