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Stream: learning: questions

Topic: Does my functor preserve left Kan extensions?

Jade Master (Dec 13 2023 at 11:05):

Hi, I have a functor
$\int : \mathsf{ISet} \to \mathsf{Set}^{\to}$
where $\mathsf{ISet}$ is the category of indexed sets $s : I \to \mathsf{Set}$ and "unnatural transformations" and where $\mathsf{Set}^\to$ is the arrow category of $\mathsf{Set}$. The functor sends an indexed set $s : I \to \mathsf{Set}$ to the function $\sum_{i \in I} s(i) \to I$ (i.e.\ a very degenerate version of the Grothendieck construction).
What I want to know is if this functor preserves left Kan extensions. I have a functor $f : T \to T'$ and I can take two different Kan extensions as in the following diagram:
image.png
My question is whether $\int \circ Lan_f M \cong Lan_f (\int \circ M)$. Is there some general characterization of when a functor preserves Kan extensions? Does anyone have some intuition about whether or not this Kan extension is preserved? Do I need some nice properties of the functor $f : T \to T'$ in order to say something meaningful?

Amar Hadzihasanovic (Dec 13 2023 at 12:49):

Maybe I am misinterpreting what $\mathsf{ISet}$ is, but there is at least one plausible definition where it is actually equivalent to $\mathsf{Set}^\rightarrow$ and $\int$ is one side of the equivalence. The inverse sends an arbitrary function $f: X \to Y$ to the $Y$-indexed set $y \mapsto f^{-1}(y)$.

Jade Master (Dec 13 2023 at 13:05):

Amar Hadzihasanovic said:

Maybe I am misinterpreting what $\mathsf{ISet}$ is, but there is at least one plausible definition where it is actually equivalent to $\mathsf{Set}^\rightarrow$ and $\int$ is one side of the equivalence. The inverse sends an arbitrary function $f: X \to Y$ to the $Y$-indexed set $y \mapsto f^{-1}(y)$.

Yes. It's definitely an equivalence with that inverse. Does being an equivalence help somehow?

Morgan Rogers (he/him) (Dec 13 2023 at 13:07):

Equivalences preserve everything.

Amar Hadzihasanovic (Dec 13 2023 at 13:18):

To be more specific,

• left adjoint functors preserve left Kan extensions;
• "adjoint functor theorems" are all about converses to this statement -- in a lot of cases, just preserving a special class of left Kan extensions, i.e. colimit cones, implies that something is a left adjoint functor;
• in fact, when a left Kan extension of the identity functor along $F$ exists, just preserving this single Kan extension is equivalent to $F$ being a left adjoint;
• equivalences are both left and right adjoint functors.

Jade Master (Dec 13 2023 at 16:05):

[Quoting…]

Right. That makes perfect sense. Thanks!