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. 



Probably easy: Why is f*:A^C'->A^C continuous and cocontinuous for any functor f:C->C'?
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](https://mathoverflow.net/questions/32791/how-is-the-right-adjoint-f-to-the-inverse-image-functor-f-described-for-f/32808#32808).


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. 


