Remark 7.29.8. Notation and assumptions as in Lemma 7.29.6. Suppose that in addition the original morphism of topoi $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ is an equivalence. Then the construction in the proof of Lemma 7.29.6 gives two functors

$\mathcal{C} \rightarrow \mathcal{C}' \leftarrow \mathcal{D}$

which are both special cocontinuous functors. Hence in this case we can actually factor the morphism of topoi as a composition

$\mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \rightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{C}') = \mathop{\mathit{Sh}}\nolimits (\mathcal{D}') \leftarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$

as in Remark 7.29.7, but with the middle morphism an identity.

There are also:

• 7 comment(s) on Section 7.29: Morphisms of topoi

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.

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 03CK. Beware of the difference between the letter 'O' and the digit '0'.