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The category of presheaves $Pre(C)$ on a small category $C$ is the category of functors $C^{op}\to Sets$. Since the category of sets is co-complete and every presheaf is a colimit of representable ones (i.e. presheaves of the form $Hom(-,c)$ for an object $c$ of $C$) $Pre(C)$ may be interpreted as the co-completion of $C$.

I wonder how $Pre(C)$ relates to $C$ if $C$ is already co-complete. More generally, if $D$ is a subcategory of $Pre(C)$ containing all the representable presheaves, how is the relation between $Pre(C)$ and $Pre(D)$?

As a special instance consider $C=\Delta$. Then $Pre(C)$ is the category $sSet$ of simplicial sets. Let $D$ be the full subcategory of finite simplicial sets, is $Pre(D)\cong sSet$?

Thank you.

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have you had a look at this related question and its excellent answer? mathoverflow.net/questions/59291/completion-of-a-category –  Chris Heunen Mar 25 '11 at 0:44

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You should think of $Pre(C)$ as a formal (i.e., free) cocompletion of $C$. Even if $C$ is cocomplete, by passing to $Pre(C)$ there will be "new" colimits that were not in $C$ (are not in the essential image of the Yoneda embedding). A simple example of this is the empty presheaf (which returns the empty set when evaluated at any object of $C$); there is no way this is represented by an object of $C$, since $\hom(-, c)$ has at least one element when evaluated at $c$.

Part of the problem is nomenclatural: normally when people use the word "completion" (as when completing a metric space, or completing an integral domain to its field of fractions), the embedding of an complete object into its completion is an isomorphism. Not so in the case of free completion or free cocompletion! There is an interesting discussion of this in the nLab; roughly speaking, the traditional sorts of completions are covered by the concept of "idempotent monad", especially when the unit of the monad is an embedding, but free cocompletion is far from idempotent.

If $C$ is cocomplete in a very nice sense (called "totally cocomplete", or just "total" for short), then even though the Yoneda embedding will certainly not be an equivalence of categories, there will be something like the next best thing to an inverse: the Yoneda embedding will have a left adjoint. (This is actually a definition of total cocompleteness.) Most of the large cocomplete categories that arise in practice as categories of structured sets, such as categories algebraic over $Set$ for instance, or $Top$, have this property.

As for the last example, $Pre(C)$ and $Pre(D)$ are not equivalent. You can add certain colimits to $C$ without changing the free cocompletion, but there are precious few. In a nutshell, $C$ and $D$ have equivalent presheaf categories if they are Morita equivalent, which in the case of categories enriched in $Set$ means that they have equivalent Cauchy completions or Karoubi envelopes -- the category you get by adjoining coequalizers of pairs $1_x, e: x \to x$ where $e$ is idempotent. (This type of completion is idempotent!) So you can enlarge $C$ by splitting idempotents and get the same presheaf category. In the case of categories enriched in $Ab$, the analogous notion of Karoubi envelope is obtained by adjoining direct sums and splitting idempotents. The general rule in $V$-enriched category theory is that you can adjoin to $C$ what are called "absolute colimits" -- colimits which are preserved by any $V$-enriched functor, and still get the same presheaf category $V^{C^{op}}$ up to equivalence (provided that idempotents split in $V$!), but that's the limit of what you can do if you want the same presheaf category.

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