Quirkiness with Kan Extensions
Given two functors and a left kan extension of along is a functor Lan along with a natural transformation universal in that for any pair there is a unique natural transformation such that as in the diagram
Our counterexample is meant to define intuition about this commutativity “up to universal natural transformation.” A properly commutating diagram may not actually be the Kan extension. In other words, if factors through by some then is not necessarily the same as
Spoiler Alert: This is exercise 1.1.3. in . If you want to find out for yourself, stop reading.
Take and arbitrary. Let be an arbitrary object and let denote the functor constant at We take e.g. is the functor constant at some singleton set Then, we can (by abuse of notation) write again for the functor constant at Indeed we have But, is not the (left) Kan extension of along
Why is this true? Because elements are in bijection with natural transformations By Yoneda’s lemma, these elements are in natural bijection with natural transformations This is exactly the universal property of left Kan extensions.
 Riehl, E. Categorical Homotopy Theory. Cambridge University Press, 2014.