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If $F: 𝒞 ⟶ 𝒟$ and $K: 𝒞 ⟶ ℰ$ are functors, where $𝒞$ is small and both $𝒟$ and $ℰ$ are cocomplete. How do the left Kan extension of $F$ along $K$ (${\rm Lan}_K(F)$) and the left Kan extension of $K$ along $F$ (${\rm Lan}_F(K)$) relate with one another? (Is it safe to say that they are adjoints for example?)

EDIT: Actually, I asked the question in general but what I had in mind was the special case where $K := y: 𝒞 ⟶ \widehat{𝒞}$ is the Yoneda embedding, so any tip to show that ${\rm Lan}_y(F) ⊣ {\rm Lan}_F(y)$ (or that ${\rm Lan}_F(y) \cong N_F := Hom(F(-), -)$) would be helpful as well.

Any insight on this would be very much appreciated!

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If $y : A \to[A°,Set]$ is the Yoneda embedding it is a general fact that, given a functor $f : A \to B$, there is the adjunction $$ Lan_yf \dashv Lan_fy $$ this is called nerve-realization paradigm.

As you can see e.g. here, asking that the two functors $F,K$ are related in this way is a rather strong request (in fact, the tentative proof in the answer contains a flaw, and to this day no convincing example, different than Yoneda extensions, has been given)

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  • $\begingroup$ Thank you! That was the special case I had in mind when I wrote the question actually (I have edited it): do you know any nice proof of $Lan_yf \dashv Lan_fy$ (avoiding if possible too much co/end fu (as in the answer you pointed out))? $\endgroup$ – Cooke4 Jul 30 '18 at 9:40
  • $\begingroup$ Or even better: a proof of ${\rm Lan}_F(y) \cong N_F := Hom(F(-), -)$ $\endgroup$ – Cooke4 Jul 30 '18 at 9:48
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    $\begingroup$ @Cooke4 The latter is easy: the category of elements of $\operatorname{Hom}(F(-),D)$ is precisely the comma category $(F\downarrow D)$ for any $D \in \mathcal D$. Hence the formula computing the pointwise left Kan extensions together with the Yoneda lemma (or mor precisely the density of the Yoneda embedding, which is a consequence) gives you the result. $\endgroup$ – Pece Jul 30 '18 at 10:20
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    $\begingroup$ You have to prove that $Nat(\hom(F,1),G)\cong Nat(y, GF)$; this can even be done directly! :-) $\endgroup$ – Fosco Loregian Jul 30 '18 at 15:10
  • $\begingroup$ @FoscoLoregian The answer of Pece leads to a pretty straightforward result, but I can't figure out how to show that $Nat(\hom(F,1),G)\cong Nat(y, GF)$, since it is supposed to be natural in $G$ if I'm not wrong: any hint? :-) $\endgroup$ – Cooke4 Aug 3 '18 at 11:04

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