Kan extensions as weighted limits

Question

In this Blog David Jaz Myers says that right Kan extensions can be expressed using weighted limits. I would really appreciate any help understanding how this is so. I do not understand the sense in which weighted limits are functorial in weights.

Answer 1

$\require{AMScd}$Let's say the types of $W,F$ are as follows:

$$ \begin{CD} \cal V @<W<< \cal C @>F>> \cal D.\end{CD}$$

If you only want to use the universal property that says $$ [{\cal C},{\cal V}](W,{\cal D}(X,F-)) \cong {\cal D}(X,\{ W,F\}) $$ then the (contravariant) functoriality in $W$ is a matter of naturality and Yoneda, similarly to what happens in the case of conical limits here.

If you allow for a representation of a weighted limit as an end,

$$ \{W,F\} \cong \int_C WC\pitchfork FC$$

where $\_\pitchfork\_$ denotes cotensor, then the functoriality in $W$ follows from functoriality of $\_\pitchfork FC$ and of the end operation $\int_{\cal C} : [{\cal C}^\text{op}\times {\cal C}, {\cal D}] \to \cal D$, because a natural transformation $W\Rightarrow U$ of weights $W,U : {\cal C}\to \cal V$ induces a wedge

$$\begin{CD} \int_C UC\pitchfork FC @. \\ @VVV @. \\ UX\pitchfork FX @>>> WX\pitchfork FX \end{CD}$$

which by the universal property of $\{W,F\}$ induces a unique $\int_C UC\pitchfork FC \to \int_C WC\pitchfork FC$.

Now, the fact that the right Kan extension of $F$ along $G$ computed in $X$ is precisely the end $$ \int_C \hom(X,GC)\pitchfork FC$$ says that $(\text{Ran}_GF)X$ is precisely the limit of $F$ weighted by the "representable" weight $\hom(X,G-)$, i.e. that $$(\text{Ran}_GF)X \cong \{\hom(X,G-),F\}.$$