The Stacks project

Lemma 4.19.8. Let $\mathcal{I}$ be an index category, i.e., a category. Assume

  1. for every pair of morphisms $a : w \to x$ and $b : w \to y$ in $\mathcal{I}$ there exist an object $z$ and morphisms $c : x \to z$ and $d : y \to z$ such that $c \circ a = d \circ b$, and

  2. for every pair of morphisms $a, b : x \to y$ there exists a morphism $c : y \to z$ such that $c \circ a = c \circ b$.

Then $\mathcal{I}$ is a (possibly empty) union of disjoint filtered index categories $\mathcal{I}_ j$.

Proof. If $\mathcal{I}$ is the empty category, then the lemma is true. Otherwise, we define a relation on objects of $\mathcal{I}$ by saying that $x \sim y$ if there exist a $z$ and morphisms $x \to z$ and $y \to z$. This is an equivalence relation by the first assumption of the lemma. Hence $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ is a disjoint union of equivalence classes. Let $\mathcal{I}_ j$ be the full subcategories corresponding to these equivalence classes. The rest is clear from the definitions. $\square$


Comments (0)

There are also:

  • 3 comment(s) on Section 4.19: Filtered colimits

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Your email address will not be published.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 002X. Beware of the difference between the letter 'O' and the digit '0'.