2
$\begingroup$

In Caramello, Theories, Sites, Toposes, I read that

Every Grothendieck topos $\mathcal E$ admits a unique (up to isomorphism) geoemtric morphism $\gamma_{\mathcal E}:\mathcal E\to \bf Set$. The direct image of $\gamma_{\mathcal E}$ is the global section functor $Hom_{\mathcal E}:\mathcal E\to \bf Set$, while the inverse image functor $\gamma^*_{\mathcal E}$ is given by $S\to\coprod_{s\in S}1_\mathcal E$.

Now I have two questions:

1) if the topos is not cocomplete, does this definition make sense? I do not think that every sheaf topos over a small site is cocomplete.

2) Why is this unique? My guess is the following: by a well-known fact about topoi, it suffices to prove that the inverse image part is unique. But since it commutes with finite limits, then it should take the terminal object to the terminal object; hence, since it commutes with colimits and $\bf Set$ is generated under colimits by its terminal object, it follows that it must be as above. Am I right?

Thank you in advance.

$\endgroup$

1 Answer 1

4
$\begingroup$

1) Every Grothendieck topos is cocomplete.

2) Yes, this explanation is correct.

$\endgroup$
4
  • $\begingroup$ Thanks. Just a clue to (1)? $\endgroup$
    – W. Rether
    Commented Dec 25, 2018 at 16:30
  • 2
    $\begingroup$ I'm assuming your definition is that a Grothendieck topos is a category of sheaves on a site? Colimits in such a category are computed just like colimits of sheaves on topological spaces: To compute a colimit, you can compute it in the presheaf category (i.e. take the colimit in Set for each object in the site) and then sheafify. $\endgroup$ Commented Dec 25, 2018 at 16:41
  • $\begingroup$ Ok, thanks! I was forgetting that we have sheafification. $\endgroup$
    – W. Rether
    Commented Dec 25, 2018 at 16:42
  • $\begingroup$ Worth noting that in (2), only needing to verify uniqueness of one of the functors is a property of adjunctions far more general than anything topos-y $\endgroup$ Commented Sep 20, 2019 at 10:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .