An Example of a non-Kan extension

Question

What are some toy example of a functor $F$ that factors through some functor $K$ along some functor $H$ where it is not nessecarily the case that $(H,id_F)$ is the left Kan extension of $F$ along $K$? I am specifically looking for an example with an algebraic flavor.

Answer 1

Let $G$ be a group and $S$ a subgroup of $G$. You can see $G$ and $S$ as categories with one object each, and the inclusion $S\hookrightarrow G$ as a functor, which we name $K$. Then a functor $H:G\to \mathbf{Vect}_\mathbb{C}$ is just a complex representation of $G$, and precomposing with $K$ is just restricting to a representation of $S$; thus the left Kan extension along $K$ is the same thing as the induced representation.

Now choose any non-zero finite-dimensional vector space $V$, and let $F:S\to \mathbf{Vect}_\mathbb{C}$ be the functor corresponding to the trivial representation of $S$ on $V$, and similarly let $H:G\to\mathbf{Vect}_\mathbb{C}$ correspond to the trivial representation of $G$ on $V$. Then $HK=F$ (because restricting the trivial representation just gives the trivial representation), but $H$ is not the left Kan extension of $F$ along $K$, since the corresponding induced representation is non-trivial.

Answer 2

It may be way too late, but I did the same exact exercise and I found a very elementary counter example.

Take $\mathbb{1}$ and $\mathbb{2}$ the categories of ordinals, i.e.

• $\operatorname{Ob} \mathbb{1} = \{1\}$, $\sharp \hom_{\mathbb{1}} (1,1) = 1$.
• $\operatorname{Ob} \mathbb{2} = \{1,2\}$, $\sharp \hom_{\mathbb{2}} (1,1) = \sharp \hom_{\mathbb{2}} (1,2) = \sharp \hom_{\mathbb{2}} (2,2)= 1$ and $\sharp \hom_{\mathbb{2}} (2,1) = 0$.

In summary, $\mathbb{1} = 1$, $\mathbb{2} = 1 \to 2$.
Now take $F=1: \mathbb{1} \to \mathbb{2}$ the constant functor ($F(1) = 1$), and $K= F = 1$.

First let us find a counterexample for the left Kan extension:
Take $H = \operatorname{id}_\mathbb{2}$, we obviously have $HK = F$, but $H$ is not the left Kan extension (in fact $\operatorname{Lan}_K F = 1: \mathbb{2} \to \mathbb{2})$.
We can see that since for $G = 1: \mathbb{2} \to \mathbb{2}$ endowed with the natural transformation $1_F$, there is no natural transformation $\alpha: H \Rightarrow G$ since we would have $\alpha_2: H(2)=2 \to 1 = G(2)$, which is impossible.

Now for the right Kan extension counter example, take $H' = 1: \mathbb{2} \to \mathbb{2}$, once again $F = HK$, but for $G' = \operatorname{id}_\mathbb{2}$ endowed with the natural transformation $1_F$, there is no natural transformation $\alpha: G' \Rightarrow H'$ since we would have $\alpha'_2: G'(2)=2 \to 1 = H'(2)$, which is impossible.
In this case we have $\operatorname{Ran}_K F = \operatorname{id}_{\mathbb{2}}$.