Definition of "classifying topos"

Question

Is there a geometric theory $T$ and a Grothendieck topos $\mathcal E$ such that (2) holds but (1) doesn't:

1. $\mathcal E$ 2-represents the 2-functor $$\mathbf{GrothTop}\to\mathbf{Cat}$$ which sends a Grothendieck topos $\mathcal E$ to the category of models of $T$ in $\mathcal E$.

2. $\mathcal E$ represents the 1-functor $$\mathrm h\mathbf{GrothTop}\to\mathrm h\mathbf{Cat}\to \mathbf{Set}$$ which can be obtained by truncating the above functor to the 1-categorical level and composing with the functor sending a category $\mathcal C$ to the set $\mathrm{Ob}(\mathcal C)/\mathord{\cong}$ of all objects of $\mathcal C$ up to isomorphism.

Conversely, (1) implies (2), right?

Answer 1

This is a bit surprising to me, but the two statements turn out to be equivalent.

A topos $\mathcal{E}$ is completely determined by the functor $$\mathbf{Geom}(-,\mathcal{E}) : \mathbf{GrothTop}^\mathrm{op} \to \mathbf{Cat}$$ (this is some kind of 2-Yoneda Lemma). But we can also look at the 1-category $\mathrm{h}\mathbf{GrothTop}$ in which we identify geometric morphisms that are isomorphic. Then the (1-categorical) Yoneda Lemma says that $\mathcal{E}$ is completely determined by the functor $$\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\, : \mathrm{h}\mathbf{GrothTop}^\mathrm{op} \to \mathbf{Class}$$ to the category of classes (we have to work with classes rather than sets because there can be a proper class of geometric morphisms up to isomorphism between two toposes).

In particular, suppose that $\mathbf{Geom}(-,\mathcal{E})/\!\cong\,\,$ agrees with the functor that sends each Grothendieck topos $\mathcal{F}$ to the collection of $T$-models in $\mathcal{F}$ up to isomorphism, for some geometric theory $T$. Then $\mathcal{E}$ is the classifying topos of $T$. To prove this, we need to use the fact that the classifying topos of a geometric theory always exists (thanks to @Mike Shulman for pointing this out).