Are left and right Kan extensions ever isomorphic? So I wonder if it is possible that a left Kan extension of a fully faithful functor $F$ along some other fully faithful functor $G$ (over a small category) agrees with (is isomorphic to) the right Kan extension of $F$ along $G$ - of course when these are defined. 
(In fact, the codomain categories of the functors I have in mind are bicomplete, so I do not worry about existence).
 A: Sure. Consider the left and right Kan extension along the terminal object $t: 1 \to \text{Set}$, applied to a functor $X: 1 \to \text{Sup}$ in the category of sup-lattices. The left Kan extension $\text{Set} \to \text{Sup}$ is the functor $\text{Set} \to \text{Sup}: A \to A \cdot X$ (the coproduct of $A$ copies of $X$), and the right Kan extension along $t: 1 \to \text{Set}$ takes $X: 1 \to \text{Sup}$ takes $A$ to $X^A$, the product of $A$ copies of $X$. (As a covariant functor in $A$, the latter functor takes a map $f: A \to B$ to the map $(\text{Ran}_t X)(f): X^A \to X^B$ which whose value at $g: A \to X$ is $B \to X: b \mapsto \bigvee_{f(a) = b} g(a)$.) The canonical transformation $A \cdot X \to X^A$ is a natural isomorphism of sup-lattices, so the right and left Kan extensions are isomorphic. 
The point is that sup-lattices are infinitary commutative monoids; we are extrapolating from binary biproducts 
$$X + X \cong X \times X$$ 
of commutative monoids to infinitary biproducts 
$$A \cdot X = \sum_A X \cong \prod_A X \cong X^A$$
of sup-lattices. 
