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There's an exercise in the Handbook of Categorical Algebra (exercise 6.7.11, volume 1) where you're supposed to show the that the following theorem:

Given three categories $\mathcal{A}, \mathcal{B}, \mathcal{C}$ with $\mathcal{A}$ small and $\mathcal{C}$ cocomplete and two functors $G:\mathcal{A} \rightarrow \mathcal{C}$ and $F:\mathcal{A}\rightarrow\mathcal{B}$, the left Kan extension of $G$ along $F$ always exists.

is a corollary of the more general adjoint functor theorem:

Consider a functor $F:\mathcal{A}\rightarrow\mathcal{B}$ with $\mathcal{A}$ Cauchy complete. The following conditions are equivalent

  1. $F$ has a left adjoint
  2. $F$ is absolutely flat and satisfies the solution set condition for every object $B$ of $\mathcal{B}$

(but not of the general adjoint functor theorem, which you will recall is the one involving preservation of limits and requires that $\mathcal{A}$ be complete)

Now, as far as I can tell, the method of proof intended is to show that the functor $F^*:\mathcal{C}^\mathcal{B}\rightarrow\mathcal{C}^\mathcal{A}$ of precomposition with $F$ satisfies the conditions for the MGAFT, since the existence of Kan extensions along $F$ for every $G$ is equivalently the existence of reflections along $F^*$ for every object $G$ of $\mathcal{C}^\mathcal{A}$.

It's easy to see that $\mathcal{C}^\mathcal{B}$ is Cauchy complete, since it is cocomplete, but I run into trouble when trying to show that $F^*$ is absolutely flat.

Like, for every functor $G \in \mathcal{C}^\mathcal{A}$ I can show that the category of elements of $\text{Nat}(G,F^*-)$ is non-empty (the colimit cocone $p:G→\Delta_L$ which must exist since $\mathcal{A}$ is small and $\mathcal{C}$ is cocomplete gives us an object $(\Delta_L,p)$) but I can't seem to find any way to show any of the other cofiltration conditions...

What am I not seeing here?

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  • $\begingroup$ If I am not confusing myself, you need completeness of $\mathcal C$ for the left and cocompleteness for the right Kan extensions. At least in the construction you want to use it will be so - left adjoints are constructed with limits and right adjoints with colimits. $\endgroup$ Commented Apr 5, 2017 at 10:23
  • $\begingroup$ well, no... if you want details, a different proof from this one can be found as theorem 3.7.2 in the same book. $\endgroup$
    – twocubes
    Commented Apr 5, 2017 at 17:26
  • $\begingroup$ At least there certainly is one construction using it the way I said - generally for a functor $F:\mathcal X\to\mathcal Y$, the value of the left adjoint to $F$ on an $Y$ is the (inverse) limit of the functor $\mathcal Y{\downarrow}F\to \mathcal X$ sending $(Y\to FX)\mapsto X$ and the value of the right adjoint on it is the colimit of $F{\downarrow}\mathcal Y\to\mathcal X$ with $(FX\to Y)\mapsto X$ $\endgroup$ Commented Apr 6, 2017 at 7:49

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