## Tag `00YW`

## 7.46. Topologies

In this section we define what a topology on a category is as defined in [SGA4]. One can develop all of the machinery of sheaves and topoi in this language. A modern exposition of this material can be found in [KS]. However, the case of most interest for algebraic geometry is the topology defined by a site on its underlying category. Thus we strongly suggest the first time reader

skip this section and all other sections of this chapter!Definition 7.46.1. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$. A

sieve $S$ on $U$is a subpresheaf $S \subset h_U$.In other words, a sieve on $U$ picks out for each object $T \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$ a subset $S(T)$ of the set of all morphisms $T \to U$. In fact, the only condition on the collection of subsets $S(T) \subset h_U(T) = \mathop{Mor}\nolimits_\mathcal{C}(T, U)$ is the following rule \begin{equation} \tag{7.46.1.1} \left. \begin{matrix} (\alpha : T \to U) \in S(T) \\ g : T' \to T \end{matrix} \right\} \Rightarrow (\alpha \circ g : T' \to U) \in S(T') \end{equation} A good mental picture to keep in mind is to think of the map $S \to h_U$ as a ''morphism from $S$ to $U$''.

Lemma 7.46.2. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$.

- The collection of sieves on $U$ is a set.
- Inclusion defines a partial ordering on this set.
- Unions and intersections of sieves are sieves.
- Given a family of morphisms $\{U_i \to U\}_{i\in I}$ of $\mathcal{C}$ with target $U$ there exists a unique smallest sieve $S$ on $U$ such that each $U_i \to U$ belongs to $S(U_i)$.
- The sieve $S = h_U$ is the maximal sieve.
- The empty subpresheaf is the minimal sieve.

Proof.By our definition of subpresheaf, the collection of all subpresheaves of a presheaf $\mathcal{F}$ is a subset of $\prod_{U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})} \mathcal{P}(\mathcal{F}(U))$. And this is a set. (Here $\mathcal{P}(A)$ denotes the powerset of $A$.) Hence the collection of sieves on $U$ is a set.The partial ordering is defined by: $S \leq S'$ if and only if $S(T) \subset S'(T)$ for all $T \to U$. Notation: $S \subset S'$.

Given a collection of sieves $S_i$, $i \in I$ on $U$ we can define $\bigcup S_i$ as the sieve with values $(\bigcup S_i)(T) = \bigcup S_i(T)$ for all $T \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$. We define the intersection $\bigcap S_i$ in the same way.

Given $\{U_i \to U\}_{i\in I}$ as in the statement, consider the morphisms of presheaves $h_{U_i} \to h_U$. We simply define $S$ as the union of the images (Definition 7.3.5) of these maps of presheaves.

The last two statements of the lemma are obvious. $\square$

Definition 7.46.3. Let $\mathcal{C}$ be a category. Given a family of morphisms $\{f_i : U_i \to U\}_{i\in I}$ of $\mathcal{C}$ with target $U$ we say the sieve $S$ on $U$ described in Lemma 7.46.2 part (4) is the

sieve on $U$ generated by the morphisms $f_i$.Definition 7.46.4. Let $\mathcal{C}$ be a category. Let $f : V \to U$ be a morphism of $\mathcal{C}$. Let $S \subset h_U$ be a sieve. We define the

pullback of $S$ by $f$to be the sieve $S \times_U V$ of $V$ defined by the rule $$ (\alpha : T \to V) \in (S \times_U V)(T) \Leftrightarrow (f \circ \alpha : T \to U) \in S(T) $$We leave it to the reader to see that this is indeed a sieve (hint: use Equation 7.46.1.1). We also sometimes call $S \times_U V$ the

base changeof $S$ by $f : V \to U$.Lemma 7.46.5. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$. Let $S$ be a sieve on $U$. If $f : V \to U$ is in $S$, then $S \times_U V = h_V$ is maximal.

Proof.Trivial from the definitions. $\square$Definition 7.46.6. Let $\mathcal{C}$ be a category. A

topology on $\mathcal{C}$is given by a rule which assigns to every $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$ a subset $J(U)$ of the set of all sieves on $U$ satisfying the following conditions

- For every morphism $f : V \to U$ in $\mathcal{C}$, and every element $S \in J(U)$ the pullback $S \times_U V$ is an element of $J(V)$.
- If $S$ and $S'$ are sieves on $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$, if $S \in J(U)$, and if for all $f \in S(V)$ the pullback $S' \times_U V$ belongs to $J(V)$, then $S'$ belongs to $J(U)$.
- For every $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$ the maximal sieve $S = h_U$ belongs to $J(U)$.

In this case, the sieves belonging to $J(U)$ are called the

covering sieves.Lemma 7.46.7. Let $\mathcal{C}$ be a category. Let $J$ be a topology on $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$.

- Finite intersections of elements of $J(U)$ are in $J(U)$.
- If $S \in J(U)$ and $S' \supset S$, then $S' \in J(U)$.

Proof.Let $S, S' \in J(U)$. Consider $S'' = S \cap S'$. For every $V \to U$ in $S(U)$ we have $$ S' \times_U V = S'' \times_U V $$ simply because $V \to U$ already is in $S$. Hence by the second axiom of the definition we see that $S'' \in J(U)$.Let $S \in J(U)$ and $S' \supset S$. For every $V \to U$ in $S(U)$ we have $S' \times_U V = h_V$ by Lemma 7.46.5. Thus $S' \times_U V \in J(V)$ by the third axiom. Hence $S' \in J(U)$ by the second axiom. $\square$

Definition 7.46.8. Let $\mathcal{C}$ be a category. Let $J$, $J'$ be two topologies on $\mathcal{C}$. We say that $J$ is

finerthan $J'$ if and only if for every object $U$ of $\mathcal{C}$ we have $J'(U) \subset J(U)$.In other words, any covering sieve of $J'$ is a covering sieve of $J$. There exists a finest topology on $\mathcal{C}$, namely that topology where any sieve is a covering sieve. This is called the

discrete topologyof $\mathcal{C}$. There also exists a coarsest topology. Namely, the topology where $J(U) = \{h_U\}$ for all objects $U$. This is called thechaoticorindiscrete topology.Lemma 7.46.9. Let $\mathcal{C}$ be a category. Let $\{J_i\}_{i\in I}$ be a set of topologies.

- The rule $J(U) = \bigcap J_i(U)$ defines a topology on $\mathcal{C}$.
- There is a coarsest topology finer than all of the topologies $J_i$.

Proof.The first part is direct from the definitions. The second follows by taking the intersection of all topologies finer than all of the $J_i$. $\square$At this point we can define without any motivation what a sheaf is.

Definition 7.46.10. Let $\mathcal{C}$ be a category endowed with a topology $J$. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{C}$. We say that $\mathcal{F}$ is a

sheafon $\mathcal{C}$ if for every $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$ and for every covering sieve $S$ of $U$ the canonical map $$ \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(h_U, \mathcal{F}) \longrightarrow \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(S, \mathcal{F}) $$ is bijective.Recall that the left hand side of the displayed formula equals $\mathcal{F}(U)$. In other words, $\mathcal{F}$ is a sheaf if and only if a section of $\mathcal{F}$ over $U$ is the same thing as a compatible collection of sections $s_{T, \alpha} \in \mathcal{F}(T)$ parametrized by $(\alpha : T \to U) \in S(T)$, and this for every covering sieve $S$ on $U$.

Lemma 7.46.11. Let $\mathcal{C}$ be a category. Let $\{ \mathcal{F}_i \}_{i\in I}$ be a collection of presheaves of sets on $\mathcal{C}$. For each $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$ denote $J(U)$ the set of sieves $S$ with the following property: For every morphism $V \to U$, the maps $$ \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(h_V, \mathcal{F}_i) \longrightarrow \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(S \times_U V, \mathcal{F}_i) $$ are bijective for all $i \in I$. Then $J$ defines a topology on $\mathcal{C}$. This topology is the finest topology in which all of the $\mathcal{F}_i$ are sheaves.

Proof.If we show that $J$ is a topology, then the last statement of the lemma immediately follows. The first and second axioms of a topology are immediately verified. Thus, assume that we have an object $U$, and sieves $S, S'$ of $U$ such that $S \in J(U)$, and for all $V \to U$ in $S(V)$ we have $S' \times_U V \in J(V)$. We have to show that $S' \in J(U)$. In other words, we have to show that for any $f : W \to U$, the maps $$ \mathcal{F}_i(W) = \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(h_W, \mathcal{F}_i) \longrightarrow \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(S' \times_U W, \mathcal{F}_i) $$ are bijective for all $i \in I$. Pick an element $i \in I$ and pick an element $\varphi \in \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(S' \times_U W, \mathcal{F}_i)$. We will construct a section $s \in \mathcal{F}_i(W)$ mapping to $\varphi$.Suppose $\alpha : V \to W$ is an element of $S \times_U W$. According to the definition of pullbacks we see that the composition $f \circ\alpha : V \to W \to U$ is in $S$. Hence $S' \times_U V$ is in $J(W)$ by assumption on the pair of sieves $S, S'$. Now we have a commutative diagram of presheaves $$ \xymatrix{ S' \times_U V \ar[r] \ar[d] & h_V \ar[d] \\ S' \times_U W \ar[r] & h_W } $$ The restriction of $\varphi$ to $S' \times_U V$ corresponds to an element $s_{V, \alpha} \in \mathcal{F}_i(V)$. This we see from the definition of $J$, and because $S' \times_U V$ is in $J(W)$. We leave it to the reader to check that the rule $(V, \alpha) \mapsto s_{V, \alpha}$ defines an element $\psi \in \mathop{Mor}\nolimits_{\textit{PSh}(\mathcal{C})}(S \times_U W, \mathcal{F}_i)$. Since $S \in J(U)$ we see immediately from the definition of $J$ that $\psi$ corresponds to an element $s$ of $\mathcal{F}_i(W)$.

We leave it to the reader to verify that the construction $\varphi \mapsto s$ is inverse to the natural map displayed above. $\square$

Definition 7.46.12. Let $\mathcal{C}$ be a category. The finest topology on $\mathcal{C}$ such that all representable presheaves are sheaves, see Lemma 7.46.11, is called the

canonical topologyof $\mathcal{C}$.

The code snippet corresponding to this tag is a part of the file `sites.tex` and is located in lines 9843–10178 (see updates for more information).

```
\section{Topologies}
\label{section-topologies}
\noindent
In this section we define what a topology on a category is as
defined in \cite{SGA4}. One can develop all of the machinery of
sheaves and topoi in this language. A modern exposition of this material
can be found in \cite{KS}. However, the case of most interest for algebraic
geometry is the topology defined by a site on its underlying category.
Thus we strongly suggest the first time reader
{\bf skip this section and all other sections of this chapter}!
\begin{definition}
\label{definition-sieve}
Let $\mathcal{C}$ be a category. Let $U \in \Ob(\mathcal{C})$.
A {\it sieve $S$ on $U$} is a subpresheaf $S \subset h_U$.
\end{definition}
\noindent
In other words, a sieve on $U$ picks out for each object
$T \in \Ob(\mathcal{C})$ a subset $S(T)$ of the set
of all morphisms $T \to U$. In fact, the only condition
on the collection of subsets
$S(T) \subset h_U(T) = \Mor_\mathcal{C}(T, U)$
is the following rule
\begin{equation}
\label{equation-property-sieve}
\left.
\begin{matrix}
(\alpha : T \to U) \in S(T) \\
g : T' \to T
\end{matrix}
\right\} \Rightarrow
(\alpha \circ g : T' \to U) \in S(T')
\end{equation}
A good mental picture to keep in mind is to think of the
map $S \to h_U$ as a ``morphism from $S$ to $U$''.
\begin{lemma}
\label{lemma-sieves-set}
Let $\mathcal{C}$ be a category. Let $U \in \Ob(\mathcal{C})$.
\begin{enumerate}
\item The collection of sieves on $U$ is a set.
\item Inclusion defines a partial ordering on this set.
\item Unions and intersections of sieves are sieves.
\item
\label{item-sieve-generated}
Given a family of morphisms $\{U_i \to U\}_{i\in I}$
of $\mathcal{C}$ with target $U$
there exists a unique smallest sieve $S$ on $U$ such that
each $U_i \to U$ belongs to $S(U_i)$.
\item The sieve $S = h_U$ is the maximal sieve.
\item The empty subpresheaf is the minimal sieve.
\end{enumerate}
\end{lemma}
\begin{proof}
By our definition of subpresheaf, the collection of
all subpresheaves of a presheaf $\mathcal{F}$ is a subset of
$\prod_{U \in \Ob(\mathcal{C})} \mathcal{P}(\mathcal{F}(U))$.
And this is a set. (Here $\mathcal{P}(A)$ denotes
the powerset of $A$.) Hence the collection of sieves on $U$
is a set.
\medskip\noindent
The partial ordering is defined by: $S \leq S'$ if and only if
$S(T) \subset S'(T)$ for all $T \to U$. Notation: $S \subset S'$.
\medskip\noindent
Given a collection of sieves $S_i$, $i \in I$ on $U$ we can
define $\bigcup S_i$ as the sieve with values
$(\bigcup S_i)(T) = \bigcup S_i(T)$ for all
$T \in \Ob(\mathcal{C})$.
We define the intersection $\bigcap S_i$ in the same way.
\medskip\noindent
Given $\{U_i \to U\}_{i\in I}$ as in the statement, consider
the morphisms of presheaves $h_{U_i} \to h_U$. We simply
define $S$ as the union of the images (Definition \ref{definition-image})
of these maps of presheaves.
\medskip\noindent
The last two statements of the lemma are obvious.
\end{proof}
\begin{definition}
\label{definition-sieve-generated}
Let $\mathcal{C}$ be a category.
Given a family of morphisms $\{f_i : U_i \to U\}_{i\in I}$
of $\mathcal{C}$ with target $U$ we say the sieve
$S$ on $U$ described in Lemma \ref{lemma-sieves-set}
part (\ref{item-sieve-generated}) is the {\it sieve on $U$
generated by the morphisms $f_i$}.
\end{definition}
\begin{definition}
\label{definition-pullback-sieve}
Let $\mathcal{C}$ be a category.
Let $f : V \to U$ be a morphism of $\mathcal{C}$.
Let $S \subset h_U$ be a sieve. We define the
{\it pullback of $S$ by $f$} to be the sieve
$S \times_U V$ of $V$ defined by the rule
$$
(\alpha : T \to V) \in (S \times_U V)(T)
\Leftrightarrow
(f \circ \alpha : T \to U) \in S(T)
$$
\end{definition}
\noindent
We leave it to the reader to see that this is indeed a sieve
(hint: use Equation \ref{equation-property-sieve}).
We also sometimes call $S \times_U V$ the {\it base change}
of $S$ by $f : V \to U$.
\begin{lemma}
\label{lemma-pullback-sieve-section}
Let $\mathcal{C}$ be a category.
Let $U \in \Ob(\mathcal{C})$.
Let $S$ be a sieve on $U$.
If $f : V \to U$ is in $S$, then
$S \times_U V = h_V$ is maximal.
\end{lemma}
\begin{proof}
Trivial from the definitions.
\end{proof}
\begin{definition}
\label{definition-topology}
Let $\mathcal{C}$ be a category. A {\it topology on $\mathcal{C}$} is given
by a rule which assigns to every $U \in \Ob(\mathcal{C})$
a subset $J(U)$ of the set of all sieves on $U$ satisfying
the following conditions
\begin{enumerate}
\item For every morphism $f : V \to U$ in $\mathcal{C}$, and
every element $S \in J(U)$ the pullback $S \times_U V$
is an element of $J(V)$.
\item If $S$ and $S'$ are sieves on $U \in \Ob(\mathcal{C})$,
if $S \in J(U)$, and if for all $f \in S(V)$ the pullback
$S' \times_U V$ belongs to $J(V)$, then $S'$ belongs to $J(U)$.
\item For every $U \in \Ob(\mathcal{C})$ the
maximal sieve $S = h_U$ belongs to $J(U)$.
\end{enumerate}
\end{definition}
\noindent
In this case, the sieves belonging to $J(U)$ are called
the {\it covering sieves}.
\begin{lemma}
\label{lemma-topology-basic}
Let $\mathcal{C}$ be a category.
Let $J$ be a topology on $\mathcal{C}$.
Let $U \in \Ob(\mathcal{C})$.
\begin{enumerate}
\item Finite intersections of elements of $J(U)$ are in $J(U)$.
\item If $S \in J(U)$ and $S' \supset S$, then $S' \in J(U)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $S, S' \in J(U)$. Consider $S'' = S \cap S'$. For every
$V \to U$ in $S(U)$ we have
$$
S' \times_U V = S'' \times_U V
$$
simply because $V \to U$ already is in $S$. Hence by the second
axiom of the definition we see that $S'' \in J(U)$.
\medskip\noindent
Let $S \in J(U)$ and $S' \supset S$. For every
$V \to U$ in $S(U)$ we have $S' \times_U V = h_V$ by
Lemma \ref{lemma-pullback-sieve-section}. Thus
$S' \times_U V \in J(V)$ by the third axiom. Hence
$S' \in J(U)$ by the second axiom.
\end{proof}
\begin{definition}
\label{definition-finer}
Let $\mathcal{C}$ be a category. Let $J$, $J'$ be
two topologies on $\mathcal{C}$. We say that $J$ is
{\it finer} than $J'$ if and only if for every object
$U$ of $\mathcal{C}$ we have $J'(U) \subset J(U)$.
\end{definition}
\noindent
In other words, any covering sieve of $J'$ is a
covering sieve of $J$. There exists a finest topology
on $\mathcal{C}$, namely that topology where any sieve
is a covering sieve. This is called the
{\it discrete topology} of $\mathcal{C}$.
There also exists a coarsest topology.
Namely, the topology where $J(U) = \{h_U\}$
for all objects $U$. This is called the
{\it chaotic} or {\it indiscrete topology}.
\begin{lemma}
\label{lemma-play-with-topologies}
Let $\mathcal{C}$ be a category.
Let $\{J_i\}_{i\in I}$ be a set of topologies.
\begin{enumerate}
\item The rule $J(U) = \bigcap J_i(U)$ defines
a topology on $\mathcal{C}$.
\item There is a coarsest topology finer than
all of the topologies $J_i$.
\end{enumerate}
\end{lemma}
\begin{proof}
The first part is direct from the definitions.
The second follows by taking the intersection
of all topologies finer than all of the $J_i$.
\end{proof}
\noindent
At this point we can define
without any motivation what a sheaf is.
\begin{definition}
\label{definition-sheaf-sets-topology}
Let $\mathcal{C}$ be a category endowed with a
topology $J$. Let $\mathcal{F}$ be a presheaf of sets
on $\mathcal{C}$.
We say that $\mathcal{F}$ is a
{\it sheaf} on $\mathcal{C}$
if for every $U \in \Ob(\mathcal{C})$ and for
every covering sieve $S$ of $U$ the canonical map
$$
\Mor_{\textit{PSh}(\mathcal{C})}(h_U, \mathcal{F})
\longrightarrow
\Mor_{\textit{PSh}(\mathcal{C})}(S, \mathcal{F})
$$
is bijective.
\end{definition}
\noindent
Recall that the left hand side of the displayed
formula equals $\mathcal{F}(U)$. In other words, $\mathcal{F}$
is a sheaf if and only if a section of $\mathcal{F}$
over $U$ is the same thing as a compatible collection of sections
$s_{T, \alpha} \in \mathcal{F}(T)$ parametrized by
$(\alpha : T \to U) \in S(T)$, and this for every covering sieve $S$
on $U$.
\begin{lemma}
\label{lemma-topology-presheaves-sheaves}
Let $\mathcal{C}$ be a category. Let $\{ \mathcal{F}_i \}_{i\in I}$ be a
collection of presheaves of sets on $\mathcal{C}$. For each
$U \in \Ob(\mathcal{C})$ denote
$J(U)$ the set of sieves $S$ with the following property:
For every morphism $V \to U$, the maps
$$
\Mor_{\textit{PSh}(\mathcal{C})}(h_V, \mathcal{F}_i)
\longrightarrow
\Mor_{\textit{PSh}(\mathcal{C})}(S \times_U V, \mathcal{F}_i)
$$
are bijective for all $i \in I$. Then $J$ defines a
topology on $\mathcal{C}$. This topology is the finest
topology in which all of the $\mathcal{F}_i$ are sheaves.
\end{lemma}
\begin{proof}
If we show that $J$ is a topology, then the last statement of
the lemma immediately follows. The first and second axioms of
a topology are immediately verified. Thus, assume that
we have an object $U$, and sieves $S, S'$ of $U$
such that $S \in J(U)$, and for all $V \to U$ in $S(V)$
we have $S' \times_U V \in J(V)$. We have to show that
$S' \in J(U)$. In other words, we have to show that for
any $f : W \to U$, the maps
$$
\mathcal{F}_i(W) =
\Mor_{\textit{PSh}(\mathcal{C})}(h_W, \mathcal{F}_i)
\longrightarrow
\Mor_{\textit{PSh}(\mathcal{C})}(S' \times_U W, \mathcal{F}_i)
$$
are bijective for all $i \in I$. Pick an element
$i \in I$ and pick an element
$\varphi \in
\Mor_{\textit{PSh}(\mathcal{C})}(S' \times_U W, \mathcal{F}_i)$.
We will construct a section $s \in \mathcal{F}_i(W)$
mapping to $\varphi$.
\medskip\noindent
Suppose $\alpha : V \to W$ is an element of $S \times_U W$.
According to the definition of pullbacks we see that
the composition $f \circ\alpha : V \to W \to U$ is in $S$. Hence
$S' \times_U V$ is in $J(W)$ by assumption on the pair
of sieves $S, S'$. Now we have a commutative diagram
of presheaves
$$
\xymatrix{
S' \times_U V \ar[r] \ar[d] & h_V \ar[d] \\
S' \times_U W \ar[r] & h_W
}
$$
The restriction of $\varphi$ to $S' \times_U V$
corresponds to an element $s_{V, \alpha} \in \mathcal{F}_i(V)$.
This we see from the definition of $J$,
and because $S' \times_U V$ is in $J(W)$.
We leave it to the reader to check
that the rule $(V, \alpha) \mapsto s_{V, \alpha}$ defines
an element
$\psi \in
\Mor_{\textit{PSh}(\mathcal{C})}(S \times_U W, \mathcal{F}_i)$.
Since $S \in J(U)$ we see immediately from the definition of $J$
that $\psi$ corresponds to an element $s$ of $\mathcal{F}_i(W)$.
\medskip\noindent
We leave it to the reader to verify that the construction
$\varphi \mapsto s$ is inverse to the natural map displayed above.
\end{proof}
\begin{definition}
\label{definition-canonical-topology}
Let $\mathcal{C}$ be a category.
The finest topology on $\mathcal{C}$ such that
all representable presheaves are sheaves, see
Lemma \ref{lemma-topology-presheaves-sheaves},
is called the {\it canonical topology} of $\mathcal{C}$.
\end{definition}
```

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