The Stacks project

Lemma 4.19.5. Let $\mathcal{I}$ be an index category, i.e., a category. Assume that for every pair of objects $x, y$ of $\mathcal{I}$ there exist an object $z$ and morphisms $x \to z$ and $y \to z$. Let $M : \mathcal{I} \to \textit{Ab}$ be a diagram of abelian groups over $\mathcal{I}$. Then the colimit of $M$ in the category of sets surjects onto the colimit of $M$ in the category of abelian groups.

Proof. Recall that the colimit in the category of sets is the quotient of the disjoint union $\coprod M_ i$ by relation, see Section 4.15. Similarly, the colimit in the category of abelian groups is a quotient of the direct sum $\bigoplus M_ i$. The assumption of the lemma means that given $i, j \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and $m \in M_ i$ and $n \in M_ j$, then we can find an object $k$ and morphisms $a : i \to k$ and $b : j \to k$. Thus $m + n$ is represented in the colimit by the element $M(a)(m) + M(b)(n)$ of $M_ k$. Thus the $\coprod M_ i$ surjects onto the colimit. $\square$


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Your email address will not be published.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 09WR. Beware of the difference between the letter 'O' and the digit '0'.