Lemma 15.12.5. Let $(A, I) = \mathop{\mathrm{colim}}\nolimits (A_ i, I_ i)$ be a filtered colimit of pairs. The functor of Lemma 15.12.1 gives $A^ h = \mathop{\mathrm{colim}}\nolimits A_ i^ h$ and $I^ h = \mathop{\mathrm{colim}}\nolimits I_ i^ h$.

**Proof.**
By Categories, Lemma 4.24.6 we see that $(A^ h, I^ h)$ is the colimit of the system $(A_ i^ h, I_ i^ h)$ in the category of henselian pairs. Thus for a henselian pair $(B, J)$ we have

Here the colimit is in the category of pairs. Since the colimit is filtered we obtain $\mathop{\mathrm{colim}}\nolimits (A_ i^ h, I_ i^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$ in the category of pairs; details omitted. Again using the colimit is filtered, this is a henselian pair (Lemma 15.11.12). Hence by the Yoneda lemma we find $(A^ h, I^ h) = (\mathop{\mathrm{colim}}\nolimits A_ i^ h, \mathop{\mathrm{colim}}\nolimits I_ i^ h)$. $\square$

## 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Your email address will not be published.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.

## Comments (9)

Comment #4896 by Laurent Moret-Bailly on

Comment #4897 by Laurent Moret-Bailly on

Comment #4901 by Johan on

Comment #4902 by Johan on

Comment #5019 by 羽山籍真 on

Comment #5020 by Laurent Moret-Bailly on

Comment #5021 by 羽山籍真 on

Comment #5023 by Johan on

Comment #5035 by slogan_bot on