5
$\begingroup$

Let $(\mathsf C,J)$ be a site. Then we have the category of sheaves $\tilde{\mathsf C}$ and the category $\tilde{\tilde{\sf C}}$ of sheaves over $\tilde{\sf C}$ (both considered with the canonical topology).

There is a functor $\epsilon: \tilde{\sf C} \to \tilde{\tilde{\sf C}}$, which is the composition $\bf ay$ of the Yoneda functor $\mathbf y_C = \operatorname{Hom}(-,C)$ and the associated sheaf functor $\mathbf a$.

I have seen it repeatedly claimed that $\epsilon$ is an equivalence, but I lack background to understand the proof in the Elephant (C2.2.7). Nonetheless, Makkai and Reyes claim in First Order Categorical Logic that "it is not hard to show Lemma 1.3.14 directly" (where 1.3.14 is the statement that $\epsilon$ is an equivalence).


So I tried to prove it directly; obviously the difficulty lies in defining the functor $R$ that forms an equivalence together with $\epsilon$. Here is my attempt:

Let $\mathcal S$ be a sheaf over $\tilde{\sf C}$ (so an object of $\tilde{\tilde{\sf C}}$); then $\mathcal S = \varinjlim_i \operatorname{Hom}(-,F_i)$ for some objects $F_i$ of $\tilde{\sf C}$. For any $F$, now, $\mathcal SF$ is an equivalence class of natural transformations $[\eta: F \to F_i]$. So now I let:

$$R\mathcal S = G, GC = \left\{\left([\eta_C: FC \to F_iC],x\right) \mid F \in \operatorname{ob}\tilde{\sf C}, x \in FC\right\}$$

and for $f:D \to C$, define $Gf: GC \to GD$ by $$Gf([\eta_C:FC\to F_iC],x) = ([\eta_D:FD \to F_iD],Ff(x))$$

So I think this will provide an equivalence, but I'm not really sure because of the multitude of supposedly obvious claims (e.g. about well-definedness, and that $G$ is a sheaf).


My question now has two parts:

  1. Is this definition going to work? What should I pay careful attention to?
  2. If not, is there any other explicit construction of the functor $R$?
$\endgroup$
  • $\begingroup$ You might want to try to show that $\epsilon$ is fully faithful and essentially surjective instead of finding an explicit inverse. The fully faithfulness seems to come from adjunction and Yoneda's lemma (I might be mistaken, I didn't write it down). It remains to show the essentially surjective part. $\endgroup$ – Pece Jul 5 '13 at 18:39
  • 2
    $\begingroup$ @Pece That might provide conceptual simplicity. But it still remains to exhibit a sheaf $F$ such that $\mathcal S \simeq \mathbf y_F$, which seems to lie at the core of the problem. $\endgroup$ – Lord_Farin Jul 5 '13 at 19:50
7
+100
$\begingroup$

For simplicity, assume $(\mathbb{C}, J)$ is a small subcanonical site. The quasi-inverse of the embedding $\mathbf{Sh}(\mathbb{C}, J) \to \mathbf{Sh}(\mathbf{Sh}(\mathbb{C}, J))$ has a very simple description: it is the functor that sends a sheaf $F : \mathbf{Sh}(\mathbb{C}, J)^\mathrm{op} \to \mathbf{Set}$ to its restriction along the embedding $\mathbb{C} \to \mathbf{Sh}(\mathbb{C}, J)$.

Indeed, suppose $F : \mathbf{Sh}(\mathbb{C}, J)^\mathrm{op} \to \mathbf{Set}$ is a sheaf. I claim $F$ is determined up to unique isomorphism by its restriction along the embedding $\mathbb{C} \to \mathbf{Sh}(\mathbb{C}, J)$. Indeed, let $X : \mathbb{C}^\mathrm{op} \to \mathbf{Set}$ be a $J$-sheaf. Then $X$ is the colimit of a canonical small diagram of representable sheaves on $(\mathbb{C}, J)$ in a canonical way. Consider the colimiting cocone on $X$: it is a universal effective epimorphic family and is therefore a covering family in the canonical topology on $\mathbf{Sh}(\mathbb{C}, J)$. Thus, $F (X)$ is indeed determined up to unique isomorphism by the restriction of $F$ to $\mathbb{C}$. We must also show that the restriction is actually a sheaf on $(\mathbb{C}, J)$; but this is true because $J$-covering sieves in $\mathbb{C}$ become universal effective epimorphic families in $\mathbf{Sh}(\mathbb{C}, J)$.

Thus we obtain a functor $\mathbf{Sh}(\mathbf{Sh}(\mathbb{C}, J)) \to \mathbf{Sh}(\mathbb{C}, J)$ that is left quasi-inverse to the embedding $\mathbf{Sh}(\mathbb{C}, J) \to \mathbf{Sh}(\mathbf{Sh}(\mathbb{C}, J))$, and the argument above shows that it is also a right quasi-inverse.

$\endgroup$
  • 1
    $\begingroup$ Thank you! That was simpler than I had thought and than I hoped for. I'll wait two more days before awarding the bounty; there may come other answers. $\endgroup$ – Lord_Farin Aug 26 '13 at 11:23
  • 1
    $\begingroup$ Well, sometimes special cases are easier to prove. The general version of the theorem (e.g. the cited one in the Elephant) is harder – I suspect there isn't an explicit description of the quasi-inverse in that case. $\endgroup$ – Zhen Lin Aug 26 '13 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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