The Stacks Project


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 $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$.

  1. The collection of sieves on $U$ is a set.
  2. Inclusion defines a partial ordering on this set.
  3. Unions and intersections of sieves are sieves.
  4. 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 $S(U_i)$.
  5. The sieve $S = h_U$ is the maximal sieve.
  6. 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 change of $S$ 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

  1. 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 $J(V)$.
  2. 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 $J(U)$.
  3. 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 $U \in \mathop{\mathrm{Ob}}\nolimits(\mathcal{C})$.

  1. Finite intersections of elements of $J(U)$ are in $J(U)$.
  2. 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 finer than $J'$ 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 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 chaotic or indiscrete topology.

Lemma 7.46.9. Let $\mathcal{C}$ be a category. Let $\{J_i\}_{i\in I}$ be a set of topologies.

  1. The rule $J(U) = \bigcap J_i(U)$ defines a topology on $\mathcal{C}$.
  2. 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 sheaf on $\mathcal{C}$ 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 topology of $\mathcal{C}$.

    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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 00YW

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?