# How is the right adjoint $f_*$ to the inverse image functor $f^*$ described for functor categories $Set^C$, $Set^D$ and $f : C \to D$

For $C,D$ small categories, and $f : C \to D$ a functor between them, there is a precomposition, or "inverse image", functor $f^* = (-) \circ f : Set^D \to Set^C$. It has a left and a right adjoint. What are their definitions, and in particular what is the right adjoint $f_*$? I couldn't find a definition in terms of functor categories, just "topological" ones.

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If I remember right, there's a good section or two on this in the Mac Lane and Moerdijk book: gen.lib.rus.ec/… Unfortunately I'm in a rush at the moment — hopefully someone else can provide chapter and verse and perhaps a digest... –  Peter LeFanu Lumsdaine Jul 21 '10 at 13:15
Pietro: I might be wrong here, but $f^*$ seems to be covariant, with $f^*(F : Set^D) = F\circ f$ and $f^*(\gamma : F \to G)_{c : C} = \gamma_{fc} : Ffc \to Gfc$. –  vincenzoml Jul 21 '10 at 14:09
The adjoints (left and right) to such a pre-composition functor are called Kan extensions. They're the subject of the last chapter of Mac Lane's "Categories for the Working Mathematician." –  Andreas Blass Jul 21 '10 at 14:09
Peter: Mac Lane and Moerdijk define $f_*$ in the "topological" way, that is (p. 68 of the copy you linked) $(f_*F)V=F(f^{−1})V$. But it is not clear to me how this definition gives us (from a presheaf $F : Set^C$) a presheaf $f_*F: Set^D$. That is: 1) What is now $F(f^{−1})(V)$? f may be non-injective on objects. 2) What is the action of $f_*F$ on arrows? $C$ may even be a discrete category, so $f^{−1}$ of a arrow in $D$ may be undefined. –  vincenzoml Jul 21 '10 at 14:23
vincenzo: sorry you're right. Not concentrated! –  Pietro Majer Jul 21 '10 at 18:37

Given a functor $f:\mathcal{C}\to\mathcal{D}$ and any complete category $\mathcal{A}$ (e.g., take $\mathcal{A}=\text{Sets}$ to get the case you are asking about), there exists a right-adjoint $f_{\*}:[\mathcal{C},\mathcal{A}]\to[\mathcal{D},\mathcal{A}]$ to the "inverse image functor" $f^{*}$ and this is given by taking right Kan extension.
Explicitly, given a functor $X:\mathcal{C}\to\mathcal{A}$, the functor $f_{*}(X):\mathcal{D}\to\mathcal{A}$ is the right Kan extension of $X$ along $f$. This can be described explicitly using the limit formula $$f_{\*}(X)(d)=\text{lim}_{d\to f(c)}X(c)$$ for $d$ an object of $\mathcal{D}$ (the action on arrows of $\mathcal{D}$ is then induced by the universal property of limits). The indexing category of the limit here is of course the comma category $(d\downarrow f)$.
When $\mathcal{A}$ is cocomplete there is a corresponding left-adjoint $f_{!}\dashv f^{*}$ which is given by taking left Kan extension along $f$. This can be explicitly described by the colimit formula dual to the limit formula given above.
Michael, thanks! This helped me understand the definition in MacLane. But reading p. 237-238 (in the second edition) that is, the definition you spell out above, and having made calculations, it seems to me that there's a typo above: the index category of the limit is $(d \downarrow f)$ and the limit is indexed over $d \to f(c)$. For the rest, a very clear explanation. –  vincenzoml Jul 22 '10 at 12:00
You're quite right about the typo. I'm so used to using this formula when dealing with presheaves $[\mathcal{C}^{\text{op}},\textbf{Sets}]$ that I mixed up the variance, doh! –  Michael A Warren Jul 22 '10 at 15:04