The Stacks project

Lemma 64.3.2. Let $S$ be a scheme. Let $a : F \to G$ be a transformation of functors $(\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. The following are equivalent

  1. $a : F \to G$ is limit preserving, and

  2. for every affine scheme $T$ over $S$ which is a limit $T = \mathop{\mathrm{lim}}\nolimits T_ i$ of a directed inverse system of affine schemes $T_ i$ over $S$ the diagram of sets

    \[ \xymatrix{ \mathop{\mathrm{colim}}\nolimits _ i F(T_ i) \ar[r] \ar[d]_ a & F(T) \ar[d]^ a \\ \mathop{\mathrm{colim}}\nolimits _ i G(T_ i) \ar[r] & G(T) } \]

    is a fibre product diagram.

Proof. Assume (1). Consider $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ as in (2). Let $(y, x_ T)$ be an element of the fibre product $\mathop{\mathrm{colim}}\nolimits _ i G(T_ i) \times _{G(T)} F(T)$. Then $y$ comes from $y_ i \in G(T_ i)$ for some $i$. Consider the functor $F_{y_ i}$ on $(\mathit{Sch}/T_ i)_{fppf}$ as in Definition 64.3.1. We see that $x_ T \in F_{y_ i}(T)$. Moreover $T = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} T_{i'}$ is a directed system of affine schemes over $T_ i$. Hence (1) implies that $x_ T$ the image of a unique element $x$ of $\mathop{\mathrm{colim}}\nolimits _{i' \geq i} F_{y_ i}(T_{i'})$. Thus $x$ is the unique element of $\mathop{\mathrm{colim}}\nolimits F(T_ i)$ which maps to the pair $(y, x_ T)$. This proves that (2) holds.

Assume (2). Let $T$ be a scheme and $y_ T \in G(T)$. We have to show that $F_{y_ T}$ is limit preserving. Let $T' = \mathop{\mathrm{lim}}\nolimits _{i \in I} T'_ i$ be an affine scheme over $T$ which is the directed limit of affine scheme $T'_ i$ over $T$. Let $x_{T'} \in F_{y_ T}$. Pick $i \in I$ which is possible as $I$ is a directed set. Denote $y_ i \in F(T'_ i)$ the image of $y_{T'}$. Then we see that $(y_ i, x_{T'})$ is an element of the fibre product $\mathop{\mathrm{colim}}\nolimits _ i G(T'_ i) \times _{G(T')} F(T')$. Hence by (2) we get a unique element $x$ of $\mathop{\mathrm{colim}}\nolimits _ i F(T'_ i)$ mapping to $(y_ i, x_{T'})$. It is clear that $x$ defines an element of $\mathop{\mathrm{colim}}\nolimits _ i F_ y(T'_ i)$ mapping to $x_{T'}$ and we win. $\square$


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