Lemma 4.19.3. Let $\mathcal{I}$ be a category. Let $\mathcal{J}$ be a full subcategory. Assume that $\mathcal{I}$ is filtered. Assume also that for any object $i$ of $\mathcal{I}$, there exists a morphism $i \to j$ to some object $j$ of $\mathcal{J}$. Then $\mathcal{J}$ is filtered and cofinal in $\mathcal{I}$.

Proof. Omitted. Pleasant exercise of the notions involved. $\square$

There are also:

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

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).