The Stacks project

Lemma 7.47.2. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 00YZ. Beware of the difference between the letter 'O' and the digit '0'.