Are Kan extensions extensions in the traditional sense?

Question

Suppose $A$ is a full subcategory of $B$. Given $F:A\to Z$, is it true that the left Kan extension agrees with $F$ on $A$, ie. for all $a\in A$, $\mathrm{Lan}_i F(a)\simeq F(a)$, where $i:A\to B$ is the inclusion?

Answer 1

Not necessarily. For a counterexample, see one of Tim Campion's counterexamples here: https://mathoverflow.net/questions/220246/what-is-the-point-of-pointwise-kan-extensions

In a wide range of cases, though, this is true: namely, when the Kan extension is pointwise. In this context, that implies $\mathrm{Lan}_iF(a)=\mathrm{colim}_{b\in i/a} F(b)$. Since $i$ is fully faithful, $i/a$ is isomorphic to $A/a$, which has the identity of $a$ as its terminal object, so this colimit is simply $F(a)$, which is what we were looking for. This is a basic example of the principle that general Kan extensions can be very perverse-it's pointwise extensions that capture all interesting and reasonable examples.

Answer 2

You might be interested in these notes I wrote a few months ago. When I say "It turns out that this definition, albeit correct, it too general" I mean precisely what is contained in Kevin's answer!