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Given any functor between two small categories, one may construct two functors between their presheaf categories. Let $f:\mathcal{A}\rightarrow\mathcal{B}$ be such a functor. Then we may extend this functor to a functor, $f_{\ast}: \hat{\mathcal{A}} \rightarrow \hat{\mathcal{B}}$ by co-continuity. On the other hand we have a functor, $f^{\ast}:\hat{\mathcal{B}}\rightarrow\hat{\mathcal{A}}$ defined by the composite, i.e., by taking $X \mapsto X \circ f^{op}$. My question is: are these two functors adjoint?

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I appologize about the terrible formatting job. – jackie boy Dec 15 '10 at 17:24
Yes, it is well-known that $f_\ast$ (under your notation; often one sees $f_!$) is left adjoint to $f^\ast$. It is saying that $f_\ast$ is the left Kan extension along $f^{op}$, and the requisite formula is a standard exercise in the yoga of the Yoneda lemma. A chatty discussion of it can be found at – Todd Trimble Dec 15 '10 at 18:06
You actually get THREE adjoint functors $f_!$, $f^*$, and $f_*$, with each adjoint to the adjacent. – David Carchedi Dec 15 '10 at 18:15

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