Let $C$ be a locally small category. Then we have the Yoneda embedding $Y : C \to \widehat{C} := [C^{op},Set]$. Recall that $C$ is called total when $Y$ has a left adjoint.

My first question is: Why aren't there any set-theoretic obstructions for such a left adjoint to exist? The category $\widehat{C}$ is not locally small, lives in a bigger universe than $C$. In some papers I've read that, for instance, algebraic categories are total. Ok, but how does the left adjoint of $Y$ looks like explicitly for $C=\mathsf{Ring}$ for instance? I'm not yet convinced that such a functor can exist ...

The second question is: What do you think of the following variant of totality (and is it already known and studied in the literature): Define $\widehat{C}$ to be the category of presheaves on $C$ which are small colimits of representable functors. Equivalently, these presheaves are cofinally small. Then $\widehat{C}$ lives in the same universe of $C$, and in fact the Yoneda embedding enjoys the universal property of the universal cococompletion of $C$ (see here).

In this setting, I can show that the category of affine schemes $\mathsf{Aff}$ is total (i.e. $\mathsf{Ring}$ is cototal), and in fact the corresponding adjunction is quite well-known: To every affine scheme $X$ we associate its functor of points $\hom(-,X)$. To every cofinally small presheaf on affine schemes $F$ we associate $\mathcal{O}(F) := \hom(F,\mathbb{A}^1)$ with component-wise ring operations (or rather we associate $\mathrm{Spec} \mathcal{O}(F)$). Note that we could also do this construction for every presheaf $F$ (and this is done in many texts, for example in J. S. Milne's script on algebraic groups), but then $\mathcal{O}(F)$ has no chance of being a small set and therefore an object of $\mathsf{Ring}$.


1 Answer 1


Totality is for example a consequence of cocompleteness and the existence of a small dense subcategory. If you have such a subcategory, with inclusion $K\colon A \rightarrow C$, say, then the functor $\widetilde{K} \colon C \rightarrow [A^{\mathrm{op}},\mathrm{Set}]$ which sends $c \in C$ to $C(K-,c)$ is fully faithful. Cocompleteness of $C$ implies that it has a left adjoint $L$. The reflection of a presheaf $F \colon C^{\mathrm{op}} \rightarrow \mathrm{Set}$ is then given by $L(F \circ K^{\mathrm{op}})$. To see this, note that for every $c \in C$, the functor $C(-,c)$ is the right Kan extension of $C(K-,c)$ along $K^{\mathrm{op}}$, which in turn follows from the fact that $C(-,c)$ is continuous and $A$ is dense. See for example

A survey of totality for enriched and ordinary categories. Cahiers de Topologie et Géométrie Différentielle Catégoriques, 27 no. 2 (1986), p. 109-132

(Section 6) for more examples and a more elaborate theorem, of which the above is a consequence.

Your second suggestion seems to me to be equivalent to cocompleteness of the category in question. Tautologically, every representable functor admits a reflection, given by the representing object. If your category is (small) cocomplete, then the presheaves which admit reflections are closed under small colimits.

  • $\begingroup$ Daniel, I am a bit suspicious about your claim that Martin's condition is equivalent to cocompleteness. Could you elaborate more on it? $\endgroup$ Jun 6, 2012 at 8:11
  • 2
    $\begingroup$ A fancier way to look at this: the category of small colimits of representables is the free cocompletion of a locally small category. It forms the object part of a pseudomonad of the Kock-Zoeberlein (or lax-idempotent) type on the category of locally small categories. Thus a locally small category is (small) cocomplete if and only if the unit (Yoneda embedding) has a left adjoint. See e.g. Section 2 of "Lex colimits" by Garner and Lack, Propositions 2.1 and 2.2. $\endgroup$ Jun 6, 2012 at 10:02
  • $\begingroup$ (here by "Yoneda embedding" I mean the corestriction of the usual Yoneda embedding to small presheaves) $\endgroup$ Jun 6, 2012 at 10:05
  • $\begingroup$ Thx, Daniel :-) $\endgroup$ Jun 6, 2012 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.