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 $S \subset h_ U$.

## 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**!

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

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

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

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

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.

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 *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

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

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

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

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 $\mathcal{C}$.

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