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Let $f:C\to C'$ be a functor, and let $A$ be a locally presentable, complete, and cocomplete category. Then according to the paper I'm reading, the pullback functor, $f^*:A^{C'}\to A^C$ (given by precomposition with $f$), admits left and right adjoints $f_!$ and $f_*$. It's clear that the proof of this fact follows from the adjoint functor theorem, so it suffices to show that $f^*$ is continuous and cocontinuous.

However, it's not clear to me how to show this fact.


Using the notation above, why is $f^*$ continuous and cocontinuous?

Sorry if this ends up being too easy.

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I think your $f^*$ should be denoted $f_*$. Then it's left adjoint is given by a Kan extension. – Martin Brandenburg Jul 24 '10 at 7:50
I don't think that's right. f*=Hom(f,A). It has an upper star because it is the image under a contravariant functor. This notation is standard in descent theory as well (cf. Stacks-GIT, for instance). – Harry Gindi Jul 24 '10 at 7:57
I just thought about the example $f_* : Sh(X) \to Sh(Y)$ for a continunous map $f : X \to Y$. – Martin Brandenburg Jul 24 '10 at 10:04
Yeah, you flipped the variance. – Harry Gindi Jul 24 '10 at 10:12
up vote 5 down vote accepted

Whenever $A$ has all small (co)limits, small (co)limits in functor categories $A^C$ are pointwise: a cone $F : I \to A^C$ is a (co)limit iff each of its “components” or “partial evaluations” $F(c) : I \to A$ is a (co)limit.

But the property of being a pointwise (co)limit is obviously preserved by composition with $f:C' \to C$. So $f^*$ is continuous and co-continuous (and you can apply the AFT).

On the other hand, as Martin says in comments, we don't really need the AFT: the adjoints are given by left and right Kan extensions, which we can write down explicitly as (co)limits, as described by Michael Warren in this recent answer.

All of this is discussed in the chapters on limits and Kan extensions in Mac Lane Categories for the working Mathematician. Also: we don't need the local presentability for any of this, and the limits and colimits halves each work independently of the other.

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