# Probably easy: Why is f*:A^C'->A^C continuous and cocontinuous for any functor f:C->C'?

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.

Question:

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

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).