As is discussed e.g. at https://ncatlab.org/nlab/show/free+cocompletion , the Presheaf category $PSh(C)=[C^{op},Set]$ over a small category $C$ can be described by an universal property, it is the unique (up to equivalence of categories) cocomplete category such that for every functor $F:C\rightarrow D$ into a cocomplete category $D$, there is a unique up to natural equivalence right exact functor $F':PSh(C) \rightarrow D$ such that $F'\circ Y =F$ with $Y$ the Yoneda embedding. One therefore also calls it the free cocompletion of $C$.

My question is the following: Since one can also view $F'$ as a functor into a cocomplete category, one also gets a essentially unique right exact $F'':PSh(PSh(C))\rightarrow D$, so the "double Presheaf" category also fulfills the described universal property and we must suppose that $PSh(PSh(C)) \simeq PSh(C)$. Analogously, showing that it fulfills the universal property, one could demonstrate that for $C$ cocomplete, $PSh(C) \simeq C$.

This looks very dubious, as if you take the terminal category $*$, then $PSh(*)\cong Set$ is locally small, but $Psh(PSh(*))\cong[Set^{op},Set]$ is obviously not, but hom-sets should be bijective under equivalences, so this makes no sense. But what is the error in the above proof? Is it size issues, since $PSh(C)$ is not a small category anymore?

  • $\begingroup$ This 13-lines single-paragraph is clearly hard to read. Consider adding some spaces. $\endgroup$
    – b00n heT
    Oct 16, 2019 at 14:49
  • $\begingroup$ (A) $PSh(PSh(C))$ is very definitely not small, so it doesn't satisfy the hypothesis of the theorem, and (B) $PSh(PSh(C))$ would be the free cocompletion of $PSh(C)$, but there's no reason to think it would be the free cocompletion of $C$ also. $\endgroup$ Oct 16, 2019 at 15:08
  • 1
    $\begingroup$ As for why a cocomplete $C$ wouldn't be its own free cocompletion, that's a better question; it would meet the criterion and is clearly not equivalent to $PSh(C)$. My suspicion is that "freeness" is being used loosely. $\endgroup$ Oct 16, 2019 at 15:38
  • $\begingroup$ What embedding of a cocomplete $C$ into itself do you think exhibits it as its own free cocompletion? The Yoneda embedding doesn't make any typing sense, so it's not clear (to me) in what sense you think it is fitting 'the same' properties as the ones you wrote. Same with the double presheaf. $\endgroup$
    – Dan Doel
    Oct 16, 2019 at 17:10
  • $\begingroup$ @Malice Vidrine That is true, but one could easily extend the theorem to higher Grothendiek universes, so I think at least at that point size is not the problem $\endgroup$ Oct 16, 2019 at 17:34

1 Answer 1


There are 2 issues at hand here : one is that you missed an important word in the property, and another one is a (non stupid) size issue.

First of all, the property says that any functor $C\to D$ with $D$ cocomplete yields a cocontinuous functor $Psh(C)\to D$.

This explains why even if $C$ is cocomplete, it's not its own cocompletion : not any functor leaving $C$ into a cocomplete category is cocontinuous ! It doesn't solve the $Psh(Psh(C))$ question, because any functor $C$ should yield a unique cocontinuous functor leaving $Psh(Psh(C))$. This is where the size issue comes in.

Often times in category theory we don't worry about size issues, but there are times where they are important, specifically when it comes to (co)limits. An example where size issues are relevant is the following : any small complete category is a preorder.

The relevance of size here is that (co)completeness is not defined with respect to all limits, but only with respect to small limits, that's why some arguments work for small categories and others not.

Here specifically, (if $C$ is nonempty) $Psh(C)$ is never (essentially) small, and so the Yoneda embedding $Psh(C) \to Psh(Psh(C)) $ (even though it could be considered, simply by passing to a bigger universe) is not dense in the usual sense : indeed in the proof that $C\to Psh(C)$ is dense uses the fact that any presheaf is a colimit of representable ones, but the colimit is indexed by a category of size comparable to $C$ - if $C$ is not small, this is not a small colimit, and so the assumptions on $D$ cannot get you anywhere (as we have never assumed that $D$ had non small colimits).

So the universal property actually does not go through for non small categories, in particular it does not work for $Psh(C)\to Psh(Psh(C))$ (although again, a priori, considering this embedding is not such a problem), and therefore you cannot apply your argument, which solves the problem.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .