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When $C$ is essentially small, the presheaf category $[C^\mathrm{op},\mathsf{Set}]$ is the free cocompletion of $C$. The presheaf category $[C^\mathrm{op},\mathsf{Set}]$ is also a topos.

When $C$ is truly large, the presheaf category $[C^\mathrm{op},\mathsf{Set}$] is not even locally small. But the full subcategory $\mathcal{P}C \subset [C^\mathrm{op},\mathsf{Set}]$, consisting of the presheaves which are small colimits of representables, is still the free cocompletion of $C$. When $C$ is small, $\mathcal{P}C$ is the whole category $[C^\mathrm{op},\mathsf{Set}]$, so it is a topos. But when $\mathcal{C}$ is large, is $\mathcal{P}C$ a topos?

Day and Lack, following Rosicky, study when $\mathcal{P}C$ is complete, and when it is cartesian closed. Certainly cartesian closedness is necessary to be a topos; I believe that a cocomplete topos is also complete, so that completeness is also necessary. But I suspect that $\mathcal{P}C$ might only have a subobject classifier if $\mathcal{C}$ is essentially small.

If it makes a difference to restrict to sheaves in some Grothendieck topology, that would be interesting to know. I suppose my motivating example is when $C$ is the category of schemes (or just affine schemes). Algebraic geometers are very interested in sheaves on this category in various topologies; I suspect most such interesting sheaves are small, but I don't know if that restriction gives one a topos to work with.

EDIT

As David Carchedi points out in the comments, I should emphasize that I'm interested in when $\mathcal{P}C$ is an elementary topos (a finitely complete, cartesian closed category with a subobject classifier). As Eric Wofsey argues in the comments, if $C$ is essentially large, then $\mathcal{P}C$ is never a Grothendieck topos (it doesn't have a small generating set).

EDIT

The weakest of observations: the Yoneda embedding $C \to \mathcal{P}C$ is still continuous, so preserves monos, and is fully faithful; so it induces injections on subobject lattices. So if $C$ is not well-powered to begin with, then $\mathcal{P}C$ is also not well-powered, so can't be a topos (given that it's locally small: look at the size of $\mathrm{Hom}(a,\Omega)$).

EDIT

It looks like the canonical way to give a general answer to this problem would be to generalize the work of Menni which characterizes when the exact completion of a finitely complete category is a topos. Taking the category of small sheaves is put in the same framework as exact completions in Mike Shulman's paper which Adeel linked to in the comments. So the way forward ought to be clear. I'm not sure how easy criteria along Menni's lines would be to check in particular cases like schemes, though.

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    $\begingroup$ According to this paper of Mike Shulman, when $\mathcal{P}C$ is finitely complete, it is an infinitary pretopos, so by Giraud's theorem it is a topos iff it admits a small generating set. $\endgroup$ – AAK Mar 30 '15 at 18:40
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    $\begingroup$ Regarding whether restricting to sheaves in a topology makes a difference, Giraud's theorem gives examples: if $C$ is a Grothendieck topos, then $Sh(C) \equiv C$. $\endgroup$ – Hurkyl Mar 30 '15 at 21:30
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    $\begingroup$ It is possible for an infinitary pretopos to be a topos without being a Grothendieck topos – take presheaves on a large group, for example. $\endgroup$ – Zhen Lin Mar 30 '15 at 22:34
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    $\begingroup$ It is possible to have a subobject classifier without being a topos. This is what seems to be happening when I think about class forcing via small sheaves. For example the Easton product gives a non-locally small infinitary pretopos with subobject classifier, but since it is not locally small the catesian closed <=> power objects construction doesn't work. $\endgroup$ – David Roberts Mar 30 '15 at 23:15
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    $\begingroup$ I keep urging Daniel Schäppi to write up what he knows about small sheaves and the fpqc site, which is not well-known, or possibly only known to experts of a certain ilk. $\endgroup$ – David Roberts Mar 30 '15 at 23:17
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The following argument shows that if $C$ is essentially large (and locally small), then the presheaf $\Omega$ of sieves on $C$ is not small. Unfortunately, as Zhen points out, this does not show that $\mathcal{P}C$ does not have a subobject classifier, since its subobject classifier ought to be the presheaf of small sieves (i.e., sieves such that the corresponding presheaf is a small presheaf). In general, not all sieves are small, and in fact it seems rather difficult to construct any small sieves (exercise: construct a locally small category in which every object has a large set of sieves, but not a single nontrivial small sieve). It thus seems unlikely to me that this argument can be turned into a proof that $\mathcal{P}C$ does not have a subobject classifier without some strong additional hypotheses. Nevertheless, I'm keeping it as an answer in case you may find it somehow helpful (and because it is a nifty argument, even thought it doesn't prove what I hoped it would!).

Suppose that $\Omega$ is a small presheaf. Then every object $a\in C$ has only a small set of sieves, and there is a small set $S$ of objects such that for every $a\in C$ and every sieve $\sigma\in\Omega(a)$, there is an object $s\in S$, a map $f:a\to s$, and a sieve $\tau\in\Omega(s)$ such that $\sigma=f^*\tau$. In fact, if any such $\tau$ exists, then there is a canonical choice, namely $\tau=f_*\sigma$.

For any $a\in C$, let $\sigma_a$ be the sieve consisting of all maps $b\to a$ that do not have a right inverse (this is the second-largest sieve on $a$). We can thus find some $s_a\in S$ and $f_a:a\to s_a$ such that $\sigma_a=f_a^*f_{a*}\sigma_a$. Let's figure out explicitly what the condition $\sigma_a=f_a^*f_{a*}\sigma_a$ means. The sieve $f_{a*}\sigma_a$ consist of all maps of the form $f_ag$ where $g:b\to a$ does not have a right inverse. The sieve $f_a^*f_{a*}\sigma_a$ then consists of all maps $h$ such that $f_ah=f_ag$ for some $g$ that does not have a right inverse. The equation $\sigma_a=f_a^*f_{a*}\sigma_a$ holds iff the latter sieve does not contain the identity $1:a\to a$, or equivalently if every $g:a\to a$ such that $f_a=f_ag$ has a right inverse.

Since $S$ is small and $C$ is essentially large, we can find an essentially large set of objects $D\subseteq C$ such that all the objects $a\in D$ have the same associated object $s_a\in S$. That is, there is a single object $s\in S$ such that for each $a\in D$, there is a map $f_a:a\to s$ such that if $g:a\to a$ is such that $f_a=f_ag$, then $g$ has a right inverse. Now consider the sieves on $s$ generated by these maps $f_a$. Since $\Omega(s)$ is a small set, there is an essentially large set of objects $E\subseteq D$ such that the $f_a$ for $a\in E$ all generate the same sieve on $s$. That is, for any $a,b\in E$, there exist maps $g:a\to b$ and $h:b\to a$ such that $f_bg=f_a$ and $f_ah=f_b$. But then $f_ahg=f_a$, so $hg$ has a right inverse, and similarly $gh$ has a right inverse. In particular, $g$ and $h$ both have right inverses, and so $a$ and $b$ are retracts of each other. But for any object $a$, there are only an essentially small set of objects that are retracts of $a$, and this contradicts the essential largeness of $E$.

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    $\begingroup$ But not every sieve is small as a presheaf? $\endgroup$ – Zhen Lin Mar 31 '15 at 8:59
  • $\begingroup$ Hrm, good point! $\endgroup$ – Eric Wofsey Mar 31 '15 at 9:02
  • $\begingroup$ Cool! As for the exercise, it's impossible (assuming by "nontrivial" functor you mean "nonrepresentable"): the empty presheaf is always small, but never representable. But the larger point is made: starting from small presheaves, it's not hard to construct other presheaves (I'm thinking of the $\sigma_a$, not $\Omega$ actually) which are natural to consider, but have no good reason to be small. $\endgroup$ – Tim Campion Mar 31 '15 at 15:38
  • $\begingroup$ By "nontrivial" I meant "not empty or equal to the sieve of all maps to an object $a$". $\endgroup$ – Eric Wofsey Mar 31 '15 at 17:58
  • $\begingroup$ Another subtlety: there might be monos in $\mathcal{P}C$ which are not (isomorphic to) sieves, although I don't know any examples. The fact that monos are sieves in presheaf categories relies on the fact that (certain) limits are computed "pointwise". There does seem to be some literature on existence of limits in $\mathcal{P}C$, e.g. this and considerations in the Mike Shulman paper referenced by Adeel, but it would take a careful reading to determine when these limits are computed pointwise. $\endgroup$ – Tim Campion Apr 11 '15 at 14:15
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As you seem to suspect, $\mathcal{P}(C)$ is rarely a topos if $C$ is large. As a simple example, if $C$ is large discrete (identified with a large set $X$) and $\mathbf{Set}$ is the category of small sets, then $\mathcal{P}(C)$ is equivalent to $\mathbf{Set}/X$, which does not have a terminal object.

You are correct that a cocomplete topos $E$ is complete. For that we would just need small products. If $\{X_i\}_{i \in I}$ is a small family of objects of $E$, then under cocompleteness there is an obvious fibered object $p: \sum_i X_i \to \Delta I$ where the codomain $\Delta I$ is the coproduct of $|I|$ copies of the terminal object $1$. The object of sections of $p$, namely $\Pi_! p$ where $!: \Delta I \to 1$ is the unique map and $\Pi_!: E/\Delta I \to E/1 \simeq E$ is right adjoint to the pullback $I^*: E \to E/\Delta I$, is the product $\prod_i X_i$ in $E$. So completeness is certainly a necessary condition for $\mathcal{P}(C)$ to be a topos.

Much of the study of small cocompletions is bound up with the study of exact completions. Just to cite one result (Lemma 3 in this paper of Rosický): if $C$ is finitely complete, then $\mathcal{P}(C)$ is equivalent to $(\Sigma C)_{ex}$ where $\Sigma C$ is the coproduct completion of $C$ and the subscript denotes exact completion. One expert on properties of exact completions, besides Mike Shulman, is Matías Menni, and so you should probably consult some of his papers for properties of exact completions, starting with his thesis. See also here.

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  • $\begingroup$ Thanks for ferreting out that implicit question about completeness (you were too polite to mention that this isn't the first time you've explained this to me)! Also for the connection to exact completions, which is a lot more concrete than the one I was talking about in my last edit. $\endgroup$ – Tim Campion Apr 2 '15 at 21:35

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