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All my functors are from commutative rings to sets. I've seen two different definitions of a local functor. In one definition we say $X$ is local if whenever $Y$ is a functor with open cover $Y_i$ the standard sequence $$\mathrm{Mor}(Y, X) \to \prod_i\mathrm{Mor}(Y_i, X) \rightrightarrows \prod_{i, j}\mathrm{Mor}(Y_i \cap Y_j, X)$$ is exact. I've also seen a functor defined as local if for all algebras $A$ and $x_1, \ldots, x_n \in A$ that generate the unit ideal the sequence $$X(A) \to \prod_iX(A_{x_i}) \rightrightarrows \prod_{i, j}X(A_{x_ix_j})$$ is exact.

In general, my question is why are these two definitions equivalent? The second definition is what you get when you plug $Y = \hom(A, -)$ and $Y_i = D(x_i) = \hom(A_{x_i}, -)$ into the first definition and use Yoneda, so specifically it's the other direction that's confusing me. Why does being able to glue maps from affine functors and this special type of open subfunctor imply you can glue maps from any functor?

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  • $\begingroup$ Surely, you want to ask for functors $Y$ representable by a scheme here (or by a ring) $\endgroup$ – user45878 Dec 14 '17 at 9:32
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There is a size issue when talking about $\mathrm{Mor}(Y,X).$ To make a meaningful answer I'll assume, without mentioning it again, that $X,Y$ are defined relative to a fixed Grothendieck universe $U.$

There are two things going on here:

  1. The distinguished opens $A_x$ are a base for the Zariski topology, so the sheaf condition for $A$ can be checked on covers by distinguished opens
  2. It suffices to check the sheaf condition for affine $Y$

For 1, consider an arbitrary cover of $Y=h_A$ by open subfunctors. This means a collection of open subfunctors $h_{A,I}(B)=\{f\in \hom(A, B)\mid f(I)B=B\}$ for $I\in\mathcal I,$ such that that the ideals $I\in\mathcal I$ generate $B.$ There is a finite sum $\sum a_ib_i=1$ with each $a_i\in I_i\in\mathcal I,$ and hence $\{h_{A,(a_i)}\}$ covers $Y.$ When the sheaf condition holds for the original cover, it holds for the refined cover, so your second condition implies your first condition restricted to affine $Y.$

For 2, given $\phi_i:Y_i\to X$ satisfying the sheaf condition, we want to show there is a unique extension to $\phi:Y\to X.$ Fix an algebra $A$ and a map $h_A\to Y.$ By definition of $Y_i,$ the pullbacks $A_i=Y_i\times_Y h_A$ are an open cover of $h_A.$ We have a composed map $A_i\to Y_i\to X$ satisfying the sheaf condition which, by your second condition and my point 1 above, glues to a unique $h_A\to X.$ Since the map $h_A\to Y$ was arbitrary, by Yoneda's lemma this process defines a function $Y(A)\to X(A)$ for each algebra $A.$ To show this is natural we need to check for each $f:h_B\to h_A$ that the following diagram commutes. $\require{AMScd}$ $$ \begin{CD} Y(A) @>Y(f)>> Y(B) \\ @V\phi_AVV @V\phi_BVV \\ X(A) @>X(f)>> X(B) \end{CD}$$

Defining $B_i=Y_i\times_Y h_B,$ precomposition with $B_i\xrightarrow{Y_i\times f} A_i$ and postcomposition by $Y_i\xrightarrow{\phi_i} X$ in either order gives a commuting diagram $$ \begin{CD} \mathrm{Mor}(A_i,Y_i) @>>> \mathrm{Mor}(B_i,Y_i) \\ @VVV @VVV \\ \mathrm{Mor}(A_i,X) @>>> \mathrm{Mor}(B_i,X) \\ \end{CD}$$

Each element of $Y(A)$ pulls back to an element of $\mathrm{Mor}(A_i,Y_i),$ which ends up as the same element of $\mathrm{Mor}(B_i,X)$ going either way round the diagram. By the sheaf condition for the cover $\{B_i\to h_B\},$ these glue to a unique element of $X(B).$ This proves that $\phi$ is a natural transformation.

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  • $\begingroup$ I have recently followed this line of thought myself: to define $\Phi\colon Y \to X$ I pick any $y \in Y(A)$ and let $\hom(A, -) \to Y$ be the map given by the Yoneda lemma that sends $\mathrm{id}_A \mapsto y$. Do the same pullback and glue procedure you do to get $\phi_y\colon\hom(A, -) \to X$ and then defined $\Phi(y) = \phi_y(\mathrm{id})$. This certainly restricts to the appropriate $Y_i \to X$ maps so all I needed was to show it's natural and that's where I got stuck. Given $f\colon A \to B$ I can't induce a map $\hom(A, A) \to \hom(B, B)$ so I can't complete the right diagram. $\endgroup$ – Jim Jan 4 '18 at 18:08
  • $\begingroup$ @Jim: I've added an argument for naturality. $\endgroup$ – Dap Jan 4 '18 at 20:47
  • $\begingroup$ I finally see it, thanks. $\endgroup$ – Jim Jan 5 '18 at 3:21

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