Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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 $\textrm{Lan}_\alpha : \textrm{Lan}_F \Rightarrow \textrm{Lan}_G$.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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