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.



(Co)complete topoi that are not Grothendieck?
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. 




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**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). 


