Example of a factorisation of functors $F = HK$ for which the Kan extension of $F$ along $K$ is not $H$.

Question

I was reading Emily Riehl's book: Categorical Homotopy theory, and I encountered exercise 1.1.3:

Exercise 1.1.3: Construct a toy example to illustrate that if $F$ factors through $K$ along some functor $H$, it is not necessarily the case that $(H,1_F)$ is the left Kan extension of $F$ along $K$.

So I did my own "toy example" but I checked online if this was a correct example but I only found more "concrete" examples e.g. with the categories of subgroups of a fixed group $G$.

Here is my question: Could you check that my example (and computations) are valid?

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Context.

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 $K= F = 1$.

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Counter examples.

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 $\operatorname{Ran}_K F = \operatorname{id}_{\mathbb{2}}$.

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Computation of left and right Kan extensions.

A bit of work maybe: there are only $3$ endofunctors of $\mathbb{2}$ which are

• $1: \mathbb{2} \to \mathbb{2}$, the constant functor.
• $2: \mathbb{2} \to \mathbb{2}$, the constant functor.
• $\operatorname{id}_\mathbb{2}$.

First let us show that $\operatorname{Lan}_K F = 1: \mathbb{2} \to \mathbb{2}$ endowed with $1_F$.
Of course we have $1K = F = 1: \mathbb{1} \to \mathbb{2}$, now for any other functor $G: \mathbb{2} \to \mathbb{2}$ endowed with $\gamma: F \to GK$, we can observe that $\gamma$ is entirely determined by $\gamma_1: F(1) = 1 \to G(1) = GK(1)$.
Now let us show that there is a unique adequate natural transformation for each "good" functor $G$.

• $G= 1$, necessarily $\gamma= 1_F$, it's quite direct.
• $G= 2$, necessarily $\gamma_1:1 \to 2$ is the only existing arrow, so the unique natural transformation from $1$ to $G$ is $\alpha: 1 \rightarrow 2$ is $\alpha_1= \alpha_2: 1 \to 2$. Of course we have $\alpha K = \gamma$.
• $G = \operatorname{id}_\mathbb{2}$, necessarily $\gamma = 1_F$, and so the unique natural transformation from $1$ to $G$ is $\alpha$ given by $\alpha_1 = 1_1$ and $\alpha_2 : 1 \to 2$. Of course we have $\alpha K = \gamma$.

Hence $\operatorname{Lan}_K F = 1$.

Now let us show that $\operatorname{Ran}_K F = \operatorname{id}_\mathbb{2}$ endowed with $1_F$.
Of course we have $\operatorname{id}_\mathbb{2}K = F = 1: \mathbb{1} \to \mathbb{2}$, now for any other functor $G': \mathbb{2} \to \mathbb{2}$ endowed with $\gamma': G'K \to F$, we can observe that $\gamma'$ is entirely determined by $\gamma_1': G'K(1) = G'(1) \to 1 = F(1)$. So $G'(1)=1$
Now let us show that there is a unique adequate natural transformation for each "good" functor $G'$.

• $G'= \operatorname{id}_\mathbb{2}$, necessarily $\gamma'= 1_F$, it's quite direct.
• $G'= 2$, is not "good" since $G(1) = 2$.
• $G' = 1: \mathbb{2} \to \mathbb{2}$, necessarily $\gamma' = 1_F$, and so the unique natural transformation from $G'$ to $\operatorname{id}_\mathbb{2}$ is $\alpha'$ given by $\alpha_1' = 1_1$ and $\alpha_2' : 1 \to 2$. Of course we have $\alpha K = \gamma'$.

Hence $\operatorname{Ran}_K F = \operatorname{id}_\mathbb{2}$ with identity as natural transformation.