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Suppose $C$ and $D$ are sites and $F$, $G:C\to D$ two functors connected by a natural transformation $\eta_c:F(c)\to G(c)$.

Suppose further that two functors $\hat F$, $\hat G:\hat C\to\hat D$ on the respective categories of presheaves are given by $\hat F(c)=F(c)$ and $\hat G(c)=G(c)$ where I abuse the notation for the Yoneda embedding.

Is there always a natural transformation $\hat\eta_X:\hat F(X)\to \hat G(X)$?

The problem is, that in the diagram $$ \begin{array}{rcccccl} \hat F(X)&=&\operatorname{colim} F(X_j)&\to& \operatorname{colim} G(X_j)&=&\hat G(X)\\ &&\downarrow &&\downarrow\\ \hat F(Y)&=&\operatorname{colim} F(Y_k)&\to& \operatorname{colim} G(Y_k)&=&\hat G(Y) \end{array} $$ for a presheaf morphism $X\to Y$ the diagrams for the colimits may be different, or am I wrong?

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Recall: given a functor $F : \mathbb{C} \to \mathbb{D}$ between small categories, there is an induced functor $F^\dagger : [\mathbb{D}^\textrm{op}, \textbf{Set}] \to [\mathbb{C}^\textrm{op}, \textbf{Set}]$, and this functor has both a left adjoint $\textrm{Lan}_F$ and a right adjoint $\textrm{Ran}_F$. Now, given a natural transformation $\alpha : F \Rightarrow G$, there is an induced natural transformation $\alpha^\dagger : G^\dagger \Rightarrow F^\dagger$ (note the direction!), given by $(\alpha^\dagger_Q)_C = Q(\alpha_C) : Q(G C) \to Q(F C)$. Consequently, if $\eta^G_P : P \to (\textrm{Lan}_G P) F$ is the component of the unit of the adjunction $\textrm{Lan}_G \dashv G^\dagger$, we can compose with $\alpha^\dagger_{\textrm{Lan}_G P}$ to get a presheaf morphism $\alpha^\dagger_{\textrm{Lan}_G P} \circ \eta^G_P : P \to (\textrm{Lan}_G P) F$, and by adjunction this corresponds to a presheaf morphism $\textrm{Lan}_F P \to \textrm{Lan}_G P$. This is all natural in $P$, so we have the desired natural transformation $\textrm{Lan}_\alpha : \textrm{Lan}_F \Rightarrow \textrm{Lan}_G$.

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Thanks, Zhen. If I have a commutative diagram of natural transformations on sites, does this also pass over to the presheaves? –  f31 Nov 5 '12 at 18:54
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Yes. Notice that $\textrm{Lan}_\alpha P : \textrm{Lan}_F P \to \textrm{Lan}_G P$ is, by definition, the unique presheaf morphism such that $F^* (\textrm{Lan}_\alpha P) \circ \eta^F_P = \alpha^\dagger_{\textrm{Lan}_G P} \circ \eta^G_P$, and so if we had a further natural transformation $\beta : G \Rightarrow H$ we could prove that $F^*(\textrm{Lan}_{\beta \bullet \alpha} P) \circ \eta^F_P = F^*(\textrm{Lan}_\beta P) \circ F^*(\textrm{Lan}_\alpha P) \circ \eta^F_P$, which suffices to prove $\textrm{Lan}_{\beta \bullet \alpha} P = (\textrm{Lan}_\beta P) \circ (\textrm{Lan}_\alpha P)$. –  Zhen Lin Nov 5 '12 at 19:50

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