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?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ I appologize about the terrible formatting job. $\endgroup$– jackie boyCommented Dec 15, 2010 at 17:24
-
4$\begingroup$ 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 ncatlab.org/nlab/show/free+cocompletion $\endgroup$– Todd TrimbleCommented Dec 15, 2010 at 18:06
-
2$\begingroup$ You actually get THREE adjoint functors $f_!$, $f^*$, and $f_*$, with each adjoint to the adjacent. $\endgroup$– David CarchediCommented Dec 15, 2010 at 18:15
-
$\begingroup$ See several previous MO questions: mathoverflow.net/questions/33181/…, mathoverflow.net/questions/32791/…, mathoverflow.net/questions/23794/…… $\endgroup$– Peter LeFanu LumsdaineCommented Dec 18, 2010 at 22:03
Add a comment
|