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Let $\cal{A},\cal{B},\cal{X}$ be categories and let $F:\cal{A}\rightarrow\cal{B}$ be a functor, then we can induce a functor $F^*:\cal{X}^\cal{B}\rightarrow\cal{X}^\cal{A}$ in the obvious manner (where $\cal{X}^\cal{A}$ denotes the functor category with objects functors from $\cal{A}$ to $\cal{X}$ and arrows natural transformations between said functors).

Question: Is it possible to induce the functor $F^*:\cal{X}^\cal{A}\rightarrow\cal{X}^\cal{B}$ in general? And if so, what is the construction?

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Please don't use the word "derive" here, it has a technical meaning. Also, your "obvious" functor goes the wrong way. (Precomposition is contravariant!) – Zhen Lin Apr 25 '12 at 12:37
You're right, I meant to say "induce". Edited. – sdf Apr 25 '12 at 12:39
I'm confused. Pre-composition with $F$ gives you a functor $F^\ast: \mathcal{X^B} \to \mathcal{X^A}$ (it sends a functor $G: \mathcal{B} \to \mathcal{X}$ to $F^\ast(G) = G \circ F : \mathcal{A} \to \mathcal{X}$ and a natural transformation $\eta$ to $\eta F$, so $(F^\ast \eta)_A = \eta_{F(A)}$. Is this what you're asking or do you want something else? (also, the direction of $F^\ast$ in the first part of the question is messed up) – t.b. Apr 25 '12 at 12:52
@sdf: You still have the functors going the wrong way. $F : \mathcal{A} \to \mathcal{B}$ induces an "obvious" functor $F^* : \mathcal{X}^\mathcal{B} \to \mathcal{X}^\mathcal{A}$ (hence the superscript star)! – Zhen Lin Apr 25 '12 at 12:52
It appears that I made a right mess of the question. Hopefully now if makes a bit more sense. Apologies. – sdf Apr 25 '12 at 13:38
up vote 1 down vote accepted

Definition. Let $F^* : \mathcal{X}^\mathcal{B} \to \mathcal{X}^\mathcal{A}$ be the functor $P \mapsto P F$. The global left Kan extension of $F$ is a functor $F_! : \mathcal{X}^\mathcal{A} \to \mathcal{X}^\mathcal{B}$ such that $F_!$ is left adjoint to $F^*$. The global right Kan extension of $F$ is a functor $F_* : \mathcal{X}^\mathcal{A} \to \mathcal{X}^\mathcal{B}$ such that $F_*$ is right adjoint to $F^*$.

What this mean in elementary terms? If $Q : \mathcal{A} \to \mathcal{X}$ is a functor, then we are looking for a functor $P : \mathcal{B} \to \mathcal{X}$ such that $P F \cong Q$. In general such a $P$ does not exist, but $F_! Q$ and $F_* Q$ are the best approximations possible, in the following sense: there are natural transformations $$\eta_Q : Q \Rightarrow (F_! Q) F \qquad \epsilon_Q : (F_* Q) F \Rightarrow Q$$ such that for any $P : \mathcal{B} \to \mathcal{X}$ and any natural transformation $\alpha : P F \Rightarrow Q$, there is a unique natural transformation $\alpha' : P \Rightarrow F_* Q$ such that $\alpha = \epsilon_Q \bullet \alpha' F$, and for any natural transformation $\beta : Q \Rightarrow P F$, there is a unique natural transformation $\beta' : F_! Q \Rightarrow P$ such that $\beta = \beta' F \bullet \eta_Q$.

Theorem. If $\mathcal{X}$ is the category of sets and $\mathcal{A}$ and $\mathcal{B}$ are both small categories, then the global left and right Kan extensions of $F$ exist.

Proof. The left Kan extension is given by the coend formula $$(F_! Q)(b) = \int^{a : \mathcal{A}} Q(a) \times \mathcal{B}(F a, b) \cong \varinjlim_{f : F a \to b} Q(a)$$ and the right Kan extension is given by the end formula $$(F_* Q)(b) = \int_{a : \mathcal{A}} Q(a)^{\mathcal{B}(b, F a)} \cong \mathcal{X}^\mathcal{A} (\mathcal{B}(b, F(-)), Q)$$ Further details can be found in [Categories for the working mathematician, Ch. X].

More generally, if $\mathcal{X}$ is a category with enough colimits (having all colimits of size $$\max \left\lbrace |{ \operatorname{ob} \mathcal{A} }|, |{ \operatorname{mor} \mathcal{A} }|, \max \left\lbrace |{ \mathcal{B}(F a, b) }| \middle| a \in \operatorname{ob} \mathcal{A}, b \in \operatorname{ob} \mathcal{B} \right\rbrace \right\rbrace$$ is sufficient), then the global left Kan extension exists, and if $\mathcal{X}$ has enough limits (having all limits of size $$\max \left\lbrace |{ \operatorname{ob} \mathcal{A} }|, |{ \operatorname{mor} \mathcal{A} }|, \max \left\lbrace |{ \mathcal{B}(b, F a) }| \middle| a \in \operatorname{ob} \mathcal{A}, b \in \operatorname{ob} \mathcal{B} \right\rbrace \right\rbrace$$ is sufficient), then the global right Kan extension exists.

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