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Is it true that a site is subcanonical if and only if the composite of its associated sheaf functor with the Yoneda embedding is fully faithful?

Supposing the topology is subcanonical we have the following isomorphisms: $$\mathsf{Hom}_{\widehat {\mathsf C}}(\mathbf{y}X,\mathbf{y}Y)\cong \mathsf{Hom}_{\widehat {\mathsf C}}(\iota\mathbf{ay}X,\iota\mathbf{ay}Y)\cong \mathsf{Hom}_{\mathsf{Sh}(\mathsf C,J)}(\mathbf{ay}X,\mathbf{ay}Y),$$ so $\mathbf{ay}$ is fully faithful.

Conversely however, I'm having a naturality issue. We have the chain of isomorphisms below, but only the right ones are natural. I was hoping to vary $X$ and use Yoneda to prove $\mathbf{y}Y\cong \iota\mathbf{ay}Y$ and conclude $\mathbf{y}Y$ is a sheaf... $$\mathsf{Hom}_\mathsf{C}(X,Y)\cong\mathsf{Hom}_{\widehat {\mathsf C}}(\mathbf{y}X,\mathbf{y}Y)\cong \mathsf{Hom}_{\mathsf{Sh}(\mathsf C,J)}(\mathbf{ay}X,\mathbf{ay}Y)\cong \mathsf{Hom}_{\widehat{\mathsf C}}(\mathbf{y}X,\iota\mathbf{ay}Y),$$

Applying Yoneda on the right again gives $\cong \iota\mathbf{ay}Y(X)$, but I don't see why this helps... Maybe the density of the Yoneda embedding is needed?

Added. Thinking some more, maybe there is no naturality problem. The isomorphisms $\mathbf{y}X\cong \iota \mathbf{ay}X$ are natural - they're components of the natural transformation $1\Rightarrow \iota \mathbf{a}$. Hence the left isomorphisms below are also natural, and therefore they all are. $$\mathsf{Hom}_{\widehat {\mathsf C}}(\mathbf{y}X,\mathbf{y}Y)\cong \mathsf{Hom}_{\widehat {\mathsf C}}(\iota\mathbf{ay}X,\iota\mathbf{ay}Y)\cong \mathsf{Hom}_{\mathsf{Sh}(\mathsf C,J)}(\mathbf{ay}X,\mathbf{ay}Y)$$ Finally, I think the isomorphisms $\mathsf{Hom}_\mathsf{C}(X,Y)\cong\mathsf{Hom}_{\widehat {\mathsf C}}(\mathbf{y}X,\mathbf{y}Y)$ are also natural - by the extended Yoneda lemma which says the evaluation functor $\mathsf C^\text{op}\times \widehat{\mathsf{C}}\longrightarrow \mathsf{Set}$ is naturally isomorphic to $\mathsf{Nat}(\mathbf{y}^\text{op}(-),-)$.

Without sweating through all the details, are my justifications for naturality correct? Is my attempt at a proof correct?

Added later. I think my handwaving is unjustified because even if we assume $\mathbf{ay}$ is fully faithful, we aren't given the isomorphisms. In particular, they may not be the components of the natural transformations I mentioned.

Added after giving up. I started looking for counterexamples, but then found:

Exercise 3.7.2 (Borceux, vol III). Let $(\mathsf C,J)$ be a site. Prove TFAE.

  1. For all $C\in \mathsf C,R\in J(C)$, the morphisms $\mathsf C(-,f),D\in \mathsf C,f\in R(D)$ constitute in $\mathsf{PSh}(\mathsf C)$ a family of morphisms which is epimorphic and effective over the representable functors;
  2. The representable functors on $\mathsf C$ are sheaves w.r.t $J$;
  3. $\mathbf{ay}$ is fully faithful.

How to solve this exercise? Even better, is there a proof for $3\implies 2$ which does not use $1$?

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  • $\begingroup$ This might help. Representable functors are by definition presheaves. The sheaffification of a presheaf, or more precisely the presheaf morphism from the representable functor to its sheafficiation, is its reflection along the inclusion of sheaves into presheaves. Subcanonical means that the reflections of representable functors are isomorphisms. I don't immediately see how being fully faithful could force the reflections to be isos. Also, you may want to think through the effective-epimorphic characterization at ncatlab.org/nlab/show/subcanonical+coverage if you want counterexamples. $\endgroup$ Commented Oct 24, 2016 at 18:44
  • $\begingroup$ @VladimirSotirov I'd given up and started looking for a counterexample, but then I found the added exercise in Borceux vol III. The exercise involves the word "effective", but I am hopeful for a simple direct proof that circumvents the characterization you mention. $\endgroup$
    – Arrow
    Commented Oct 24, 2016 at 19:04
  • $\begingroup$ Fix $c\in C$. Consider the class $S$ of presheaves $P$ such that the canonical map $Hom(P,yc)\to Hom(P,ayc)$ is an iso. You've shown that $S$ contains representable presheaves. It remains to observe that $S$ is closed under colimits. $\endgroup$ Commented Nov 11, 2016 at 20:38
  • $\begingroup$ @MarcHoyois sorry, could you explain why that is true? $\endgroup$
    – Arrow
    Commented Nov 20, 2016 at 10:59

1 Answer 1

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You already proved that:

$$ Hom_C(X,Y) \simeq (ay Y)( X)$$

But $Hom_C(X,Y)$ is by definition $(y Y)(X)$, and all those isomorphism are functorial in $X$ so you have proved that $(y Y)( X) \simeq ay Y(X)$ which proves that $yY $ is a sheaf.

You can call that the density of the Yoneda embeddings if you want.

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  • $\begingroup$ Sorry, I don't follow. In the direction I'm stuck on, we're given isomorphisms $\mathsf{Hom}(X,Y)\cong \mathbf{ay}Y(X)$ but do not know they're natural/functorial. How do you know they are? $\endgroup$
    – Arrow
    Commented Nov 20, 2016 at 11:15
  • $\begingroup$ You obtained this isomorphism by a chain of several isomorphisms and they are all functorial. For which of them do you don't this the functoriality ? $\endgroup$ Commented Nov 20, 2016 at 14:03

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