I'm struggling to reconcile two definitions of the left Kan extension.
Suppose:
- $C$ is a discrete category with two objects $c_1,c_2$
- $D$ is a category with two objects $d_1,d_2$ and one non-identity morphism $f: d_2 \rightarrow d_1$
- $F: C \rightarrow Set$ is a functor that maps $F(c_1) = \{x_1\}, F(c_2) = \{y_1,y_2\}$
- $K: C \rightarrow D$ is a functor that maps $K(c_1) = d_1, K(c_2) = d_2$
I'm looking to find the left Kan extension of $F$ along $K$. From Riehl's Category Theory with Context we can use the following definition since $Set$ is co-complete
To evaluate $Lan_KF(d_1)$ we can see that the category $(K \downarrow d_1)$ has two objects: $$(c_1, id_{d_1}: d_1 \rightarrow d_1), (c_2, f: d_2 \rightarrow d_1)$$ Since $C$ is discrete there are no morphisms between these objects. Therefore this colimit is simply the coproduct of $F(c_1)$ and $F(c_2)$, which suggests that $Lan_KF(d_1) = F(c_1) + F(c_2) = \{x_1, y_1, y_2\}$.
However, I don't see how this makes sense by the following definition:
It seems like we can define a functor $M: D \rightarrow Set$ \begin{align*} &M(d_1) = \{x_1\}\\ &M(d_2) = \{y_1, y_2\}\\ &M(f)(y_1) = x_1, M(f)(y_2) = x_1 \end{align*} and identity natural transformation $\gamma: F \rightarrow (M \circ K)$ since: \begin{align*} &(M \circ K)(c_1) = F(c_1) = \{x_1\}\\ &(M \circ K)(c_2) = F(c_2) = \{y_1, y_2\} \end{align*}
How is $M$ not the left Kan extension? What am I missing?
As Zhen Lin points out, your definition is not initial. Consider natural transformations $φ : M → \mathsf{Lan}_K F$, where the latter is given by the point-wise definition. This requires:
$$ φ_{d_1} : \{x_1\} → \{x_1,y_1,y_2\} \\ φ_{d_2} : \{y_1,y_2\} → \{y_1,y_2\} $$
such that:
$$ φ_{d_1} \circ M(f) = \mathsf{Lan}_K F(f) \circ φ_{d_2} $$
noting that $\mathsf{Lan}_KF(f)$ is effectively the inclusion of $\{y_1,y_2\}$ into $\{x_1,y_1,y_2\}$. The left hand side of the equation is just a constant function yielding whatever $φ_{d_1}$ picks out. Because of the behavior of $\mathsf{Lan}_KF$, the constant must be either $y_1$ or $y_2$. And $φ_{d_2}$ must also be a constant map to this value. However, this means there are two valid definitions of $φ$, so $M$ is not initial.