Kan extensions and the yoneda embedding. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:38:35Z http://mathoverflow.net/feeds/question/23794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embedding Kan extensions and the yoneda embedding. Garlef Wegart 2010-05-06T22:33:06Z 2010-11-18T20:48:09Z <p>[Edit: For a category $C$ let $C^\wedge$ denote the category of presheaves on $C$.]</p> <p>Let $f:C\to D$ be a functor. Precomposition with $f^{op}$ induces a functor $f^\wedge:D^\wedge \to C^\wedge$. This functor has both a left- and a right adjoint, called <em>left-</em> and <em>right kan extension</em>:</p> <p>$f_\wedge \dashv f^\wedge \dashv f_+$.</p> <p>Now for $c\in C$ we get $D^\wedge(D(-,fc),Y)=Y(fc)=f^\wedge Y(c)=C^\wedge(C(-,c),f^\wedge Y)$. This gives us the restriction of $f_\wedge$ to $C$ along the yoneda embedding: It is $f$ (composed with the yoneda embedding).</p> <p>Now here's my question:</p> <blockquote> <p>What is the restriction of $f_+$ to $C$ along the yoneda embedding?</p> </blockquote> <p>It seems not to agree with $f$ but:</p> <blockquote> <p>Is there a nice connection between $f_+C(-,c)$ and $D(-,fc)$?</p> </blockquote> http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embedding/23802#23802 Answer by Martin Brandenburg for Kan extensions and the yoneda embedding. Martin Brandenburg 2010-05-07T01:03:20Z 2010-05-07T01:11:51Z <p>For all $Z \in C^\wedge, Y \in D^\wedge$, we have $C^\wedge(f^\wedge Y,Z)=D^\wedge(Y,f_+ Z)$. If we put $Y = D(-,d), Z = C(-,c)$, we get</p> <p>$(f_+ C(-,c))(d) = C^\wedge(f^\wedge D(-,d),C(-,c)) = C^\wedge(D(f-,d),C(-,c))$</p> <p>There seems to be no connection between $f_+ C(-,c)$ and $D(-,fc)$ (only when $f$ is an equivalence). For example,</p> <p>$D^\wedge(D(-,fc),f_+ C(-,c)) = C^\wedge(f^\wedge D(-,fc),C(-,c)) = C^\wedge(D(f-,fc),C(-,c))$</p> <p>and it is possible to construct an example where there is no natural transformation $D(f-,fc) \to C(-,c)$ at all. For example if $D(fx,fc)$ is nonempty, but $C(x,c)$ is empty. Take $C^{op}=D=Set, f = Hom(-,2), x = 0, c = 1$.</p> http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embedding/23843#23843 Answer by David Carchedi for Kan extensions and the yoneda embedding. David Carchedi 2010-05-07T10:33:35Z 2010-05-07T10:33:35Z <p>As Martin stated, the answer is no in general. However, the answer is yes if f is full and faithful and everything in $D$ can be written as a colimit of something in $C$ (i.e. you want that $D \to Set^{C^{op}}$ via $d \mapsto Hom(f(blank),d)$ to be fully faithful as well).</p> <p>In this case, $f_+$ (which is classicaly denoted $f_*$) agrees with f on representables. An example of such a situation is when $f$ is the inclusion of compact Hausdorff spaces into compactly generated Hausdorff spaces.</p> http://mathoverflow.net/questions/23794/kan-extensions-and-the-yoneda-embedding/46529#46529 Answer by Buschi Sergio for Kan extensions and the yoneda embedding. Buschi Sergio 2010-11-18T19:17:08Z 2010-11-18T20:48:09Z <p>Let $U: \mathscr{A} \to \mathscr{C}$. If $\mathscr{E}$ has (enought large) limits resp. colimits then the functor $U^*: CAT(\mathscr{C}^{op}, \mathscr{E} ) \to CAT(\mathscr{A}^{op}, \mathscr{E} ): P \mapsto P\circ U^{op}$ has a left adjoin $U_!=Lan_{ U^{op}}$ (puntual Kan extention) resp. a right adjoint $U_*=Ran_{ U^{op} }$ and $U_! \dashv U^* \dashv U_*$ with $U_!(P)(X):= {\underrightarrow{lim}} (P\circ \pi^{op}: (X\downarrow U)^{op}\to \mathscr{A} ^{op}\to \mathscr{E})$</p> <p>$U_*(P)(X):={\underleftarrow{lim}} (P\circ \pi^{op}: (U\downarrow X)^{op}\to \mathscr{E}$ ). </p> <p>Let $\mathscr{E} =Set$ and $\mathscr{A}, \mathscr{B}$ small (we can have more general conditions for the existence of puntual Kan extentions) , we have </p> <p>$U_!(P)$= $Lan_{ h_-} (h_U)(P)$ = </p> <p>$\underrightarrow{lim}$$<em>{(A, a)\in \mathscr{A} \downarrow P } h</em>{ U(A)}, </p> <p>U_*(P)(X) =\mathscr{A} ^>(h^U_X, P) </p> <p>indeed: </p> <p>(\underrightarrow{lim}$$<em>{(A, a)}$ $h</em>{U(A)}, Q)\cong$, </p> <p>${\underleftarrow{lim}}_{(A, a)} QU(A)\cong$</p> <p>$({\underrightarrow{lim}}_{(A,a)} h_A, Q\circ U)$ ; </p> <p>$(Q\circ U, P) \cong ({\underrightarrow{lim}}_{(X, x)\in \mathscr{C}\downarrow Q } h^U_X , P) \cong {\underleftarrow{lim}}_{(X, x)} (h_X, \mathscr{A} ^>(h^U_-, P)) \cong (Q(-), \mathscr{A}^>(h^U_-, P))$.</p> <p>Then (answere for you): $U_∗(h_Y)(X)=(h^U_X,h_Y)$.</p>