Timeline for Functors in Isbell duality exchange $f^*a$ and $f_*a$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2018 at 14:20 | comment | added | fosco | the question below is also adressed to you. Thanks for your help and patience! | |
| Jun 14, 2018 at 13:54 | comment | added | fosco | Now I see what was going wrong and your counterexample doesn't collide any more with my understanding of Isbell adjunction :-) ah, duality: the most underrated phenomenon in category theory. | |
| Jun 14, 2018 at 11:14 | comment | added | Noam Zeilberger | (Remember that taking opposite categories defines an operation on Cat which reverses the direction of the 2-cells.) | |
| Jun 14, 2018 at 11:06 | comment | added | Noam Zeilberger | I think that the computation you are doing actually corresponds to the left Kan extension of $y^{\sharp\circ} : A^\circ \to [A,Set]$ along $f^\circ : A^\circ \to B^\circ$. This gives you $B(1,f)^\circ : B^\circ \to [A,Set]$ rather than $B(1,f)$, and taking opposites $B(1,f)^{\circ\circ} = B(1,f)$ turns the left Kan extension into a right Kan extension. | |
| Jun 14, 2018 at 10:10 | comment | added | fosco | The second integral depends on $a$ contravariantly in both variables, this doesn't allow you to compute the coend. | |
| Jun 14, 2018 at 10:00 | comment | added | Noam Zeilberger | for the first formula, since the Kan extension is along $f : A \to B$, why do you have $\int^a y^\sharp(a) \times B^\circ(fa,b)$ rather than $\int^a y^\sharp(a) \times B(fa,b)$? | |
| Jun 14, 2018 at 9:39 | comment | added | fosco | I am sorry to continue pushing this apparently boring point, but if I try to write Kan formula for $Lan_f(y^\sharp)(b)(a')$ it comes to me that $$ \left(\int^a y^\sharp(a)\times B°(fa,b)\right)(a')\cong \int^a \hom(a,a')\times B(b,fa)=B(b,fa') $$ whereas the dual formula for $Ran_f(y^\sharp)(b)(a')$ yields $$ \left(\int_a y^\sharp(a)^{B°(b,fa)}\right)(a')\cong Nat(B(f-,b), A(-,a')) = {\cal O}(f^*(b))(a') $$ This is compatible with Todd Trimble's answer. | |
| Jun 14, 2018 at 9:08 | comment | added | Noam Zeilberger | I was confused as well, but since $[A,Set]^\circ = [A^\circ,Set^\circ]$, the point is that a natural transformation $G \Rightarrow H : B \to [A,Set]^\circ$ corresponds to a family of maps $G(b)(a) \to H(b)(a)$ in $Set^\circ$, hence a family of functions $H(b)(a) \to G(b)(a)$. Then $B(1,f)$ is a right Kan extension rather than a left Kan extension, with the counit $\epsilon : B(1,f) \circ f \Rightarrow y^\sharp$ corresponding to the family of functions $A(a',a) \to B(fa',fa)$. | |
| Jun 14, 2018 at 8:32 | vote | accept | fosco | ||
| Jun 14, 2018 at 8:32 | comment | added | fosco | Isbell adjunction is contravariant, I thought this took care of the fact you point out. But maybe I've been tricked into believing this by an incorrect number of dualizations? I agree with your counterexample, but I need to understand more what's going on. So: I'm surely accepting the answer, but can you please elaborate on $B(1,f)$ being a right and non-left extension? (perhaps you prefer to discuss the topic in chat?) | |
| Jun 14, 2018 at 7:51 | history | edited | Noam Zeilberger | CC BY-SA 4.0 |
flip upper/lower
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| Jun 14, 2018 at 7:26 | comment | added | Noam Zeilberger | @FoscoLoregian Okay, I've thought about this again (sleep is good!), and I'm pretty sure my original answer and counterexample were correct. The reason why your argument above doesn't work is simply that $f_*$ is not the left Kan extension of $y^\sharp$ along $F$, it is the right Kan extension. | |
| Jun 13, 2018 at 10:43 | history | edited | Noam Zeilberger | CC BY-SA 4.0 |
typesetting bug
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| Jun 13, 2018 at 10:11 | comment | added | Noam Zeilberger | I must be missing something. When $A = 1$, the Isbell adjunction reduces to the contravariant adjunction on Set induced by exponentiation into the singleton set. I don't see how composing with $1^{(-)}$ has a hope of transporting $f^*$ to $f_*$, or vice versa, unless $f : 1 \to B$ is an initial/terminal object. | |
| Jun 13, 2018 at 9:18 | comment | added | fosco | And yet it seems to me that ${\cal O}$ must send $Lan_fy=f^*$ to $Lan_fy^\sharp=f_*$ since it is a left adjoint: $$ {\cal O} \circ B(f,1) = {\cal O}\circ Lan_f y_A \cong Lan_f ({\cal O} \circ y_A) \cong Lan_f y_A^\sharp = B(1,f) $$ where $y^\sharp : A \to [A,Set]°$... | |
| Jun 13, 2018 at 9:11 | history | edited | Noam Zeilberger | CC BY-SA 4.0 |
consistent notation for hom
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| Jun 13, 2018 at 8:59 | history | answered | Noam Zeilberger | CC BY-SA 4.0 |