# Why "cocomplete topos"?

This is a quote from Mac Lane and Moerdijk's Sheaves in Geometry and Logic:

A classifying topos for $$T$$-models is then a topos $$\mathcal B(T)$$ over $$\mathbf{Sets}$$ with the property that for every cocomplete topos $$\mathcal E$$ there is an equivalence of categories $$c_\mathcal E\colon\mathrm{Mod}(\mathcal E, T)\to \mathrm{Hom}(\mathcal E, \mathcal B(T))$$ which is natural in $$\mathcal E$$.

Why do they say cocomplete topos and not Grothendieck topos?

It is well-known that a Grothendieck topos is the same as a cocomplete elementary topos with a small generating set. So it seems they want to emphasize that they don't assume the small generating set. But then, why do they need the axioms for an elementary topos? The subobject classifier isn't needed when interpreting geometric theories $$T$$.

Note that a category satisfying all Giraud axioms except the small generating set need not have a subobject classifier, i.e., need not be an elementary topos!

When working with classifying topoi I think there are two choices:

1. One allows $$T$$ to be large. Then the "classifying categories" one considers should be categories satisfying all Giraud axioms except the small generating set.

2. One assumes $$T$$ to be small. In this case the classifying category has a small set of generators, hence it is elementary by accident.

In both cases one doesn't get exactly cocomplete elementary toposes!

• Thanks! "But it works, because if you have a classifying topos that works for all infinitary pretoposes then it also works for all cocomplete toposes in particular." What about the converse: suppose that for all cocomplete elementary toposes $\mathcal E$, $\mathrm{Mod}(\mathcal E, T)\cong \mathrm{Hom}(\mathcal E, \mathcal F)$ (natural in $\mathcal E$). Why does it follow that for all infinitary pretoposes $\mathcal E$, $\mathrm{Mod}(\mathcal E, T)\cong \mathrm{Hom}(\mathcal E, \mathcal F)$ (natural in $\mathcal E$)? Here, $\mathcal F$ is a candidate for the classifying topos. Feb 5 at 16:24