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Recall that an elementary topos is a cartesian closed category with finite limits and a subobject classifier. A Grothendieck topos is a category equivalent to the category of sheaves on a site.

Are there examples of (co)complete elementary topoi that are not Grothendieck? On the "Cocomplete" side of things, such a category cannot be accessible, since a locally presentable elementary topos is automatically Grothendieck.

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A classical example is $G$-$Set$ for a large group $G$. That this is a cocomplete elementary topos is not hard to see. Limits and colimits are formed at the underlying set level, and exponentials $Y^X$ are formed as usual as the set of functions $f: X \to Y$ with the $G$-action $(g, f) \mapsto g f$ defined by $g f: x \mapsto g f(g^{-1} x)$. The subobject classifier is the 2-element set with trivial $G$-action (again, as usual). Thus $E = G$-$Set$ is a cocomplete (elementary) topos.

It remains to see $E$ is not Grothendieck. If it were, then every continuous functor such as the underlying-set functor $U: E \to Set$ would have a left adjoint $F$, by the special adjoint functor theorem (the hypotheses for the SAFT are satisfied: a Grothendieck $E$ is well-powered and complete and has a cogenerator $\Omega^c$ where $c$ is the coproduct of associated sheaves of objects in a small site presentation). In particular, we would have $\hom_E(F(1), -) \cong U$, but the only candidate for the representing object $F(1)$ would be $G$, which is ruled out by largeness.


Edit: Adam Epstein pointed out in a comment that the intuition in the last sentence, regarding the only candidate for the representing object, is not correct for some $G$ such as a large simple group. The following addendum patches up this oversight.

Let $G$ be a large free group, say the union of the diagram of free groups $F(\alpha)$ obtained by applying the free group functor to von Neumann cardinals $\alpha$ and initial segment inclusions between them. Then the topos $Set^G$ of $G$-sets is not Grothendieck.

Supposing it is, then the underlying-set functor $U: Set^G \to Set$ is representable, by the special adjoint functor theorem. Hence $U \cong \hom(X, -)$ for some $G$-set $X$. We will show $U$ has a proper class of non-isomorphic representable subfunctors, i.e., the object $X$ has a proper class of quotients, which is impossible in a Grothendieck topos (or even in a locally small topos -- quotients of $X$ are in bijection with equivalence relations on $X$, forming a subcollection of a hom-set $[X \times X, \Omega]$).

To begin with, for the class of canonical inclusions $i_\alpha: F(\alpha) \to G$ we may uniformly exhibit retractions $r_\alpha: G \to F(\alpha)$ (retractions in the sense of group homomorphisms). We may then view $F(\alpha)$ as a $G$-set with action $G \times F_\alpha \to F_\alpha$ taking $(g, x)$ to $r_\alpha(g) \cdot x$, and the representable functor $\hom(F(\alpha), -)$ is naturally a subfunctor of $U$. Indeed, a map $f: F(\alpha) \to A$ is uniquely determined by the value $a = f(1)$ at the identity element $1 \in F(\alpha)$, as the intertwiner condition yields $f(r_\alpha(g)) = g a$, whence $f(x) = i_\alpha(x)a$ for general $x \in F(\alpha)$. Denoting the idempotent map $i_\alpha r_\alpha: G \to G$ by $p_\alpha$, the subfunctor inclusion is given componentwise by

$$\hom(F(\alpha), A) \cong \{a \in A: \forall_{g \in G}\; g a = p_\alpha(g) a\} \hookrightarrow U(A)$$

and all these subfunctors are non-isomorphic, for the simple crude reason that the $F(\alpha)$ have generally different cardinalities as sets (as soon as $\alpha$ is in the uncountable range).

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    $\begingroup$ Incidentally, this $G$-$Set$ is complete and well-powered (clearly). So the obstruction to using SAFT is solely in the lack of a cogenerator. Also, as an aside: cocomplete toposes $E$ are automatically complete (continued next comment). $\endgroup$
    – Todd Trimble
    Dec 29, 2016 at 23:44
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    $\begingroup$ For example, let's construct the product of a family $\{B_i\}_{i \in I}$. By cocompleteness, we have a map in $E$: $$p: \sum_{i \in I} B_i \to \Delta I,$$ where $\Delta I = \sum_{i \in I} 1$, sending the summand $B_i$ to $i$ (the $i^{th}$ copy of $1$). Then $$\prod_{i \in I} B_i = \Pi_! p$$ is the product, where $!$ is the unique map $\Delta I \to 1$. Indeed, maps $X \to \Pi_! p$ are in natural bijection with maps $!^\ast X \to p$ in $E/\Delta(I) \simeq E^I$ (this equivalence holds by infinite extensivity), i.e., with an $I$-indexed family of maps $X \to B_i$, as required. $\endgroup$
    – Todd Trimble
    Dec 29, 2016 at 23:44
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    $\begingroup$ Another example is the topos obtained by gluing two Grothendieck toposes along an inaccessible left exact functor. Unfortunately one can't make this very explicit, since such functors only exist under hypotheses involving large cardinals. $\endgroup$ Dec 30, 2016 at 11:34
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    $\begingroup$ @ToddTrimble there's somewhere in the Elephant that he lists known examples of unbounded geometric morphisms to Set, which include all cocomplete locally small toposes. One is an inaccessible gluing, one is actions of a large group (or more generally presheaves on a large category with essentially small slices), and one is uniformly continuous actions of a topological group (which is not cocomplete). $\endgroup$ Dec 31, 2016 at 21:27
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    $\begingroup$ @ToddTrimble Isn't it more than that? Those examples I gave are group objects in the category of classes and class functions. Yet (by simplicity) any action by one of these large groups $G$ on a set is trivial, hence $G$-$Set$ is equivalent (indeed isomorphic) to $Set$, whence a Grothendieck topos. $\endgroup$ Jun 7, 2018 at 11:07

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