# dense generators and left Kan extensions

Here is an exercise from Borceux, Handbook of Categorical Algebra I, p 174:

Consider a category $\mathfrak{C}$, a family $(G_i)_{i \in I}$ of objects of $\mathfrak{C}$ and the corresponding full subcategory $\mathcal{G} \subseteq \mathfrak{C}$. Prove that $(G_i)_{i \in I}$ is a dense family of generators if and only if the left Kan extension of $\mathcal{G} \subseteq \mathfrak{C}$ along $\mathcal{G} \subseteq \mathfrak{C}$ is the identity on $\mathfrak{C}$

I think i can prove the "only if" part, but for the "if" part i need to take homset indexed products in $\mathfrak{C}$ (e.g. by assuming $\mathfrak{C}$ is locally small and has all products), my effort reproduced below.

I have two questions: Is the result really true in general, if so why? And if the result is not true in general then how would you go about finding a counterexample?

The effort (with details proving naturality etc ommitted for brevity) is as follows:

Let $H : \mathcal{G} \hookrightarrow \mathfrak{C}$ be the injecting functor. Say the Kan extension of $H$ along $H$ is the identity. Since $H$ is full and faithful we can take the corresponding natural transformation $H \rightarrow H$ as the identity.

Take $C$ in $\mathfrak{C}$ and let $\Gamma_C : (\mathcal{G} \downarrow C) \rightarrow \mathfrak{C}$ be the canonical projection, and $\gamma_C : \Gamma_C \rightarrow \Delta C$ be the canonical cocone. Let $\theta : \Gamma_C \rightarrow \Delta D$ be any other cocone. The goal is to show that $\hat{\theta} : C \rightarrow D$ exists uniquely such that $\theta = \Delta \hat{\theta} \cdot \gamma_C$

We set the "inputs" to the Kan extension situation as follows: Let $K : \mathfrak{C} \rightarrow \mathfrak{C}$ be the functor that takes each object $X$ to $\prod_{f : X \rightarrow C} D$ and arrows $g : X \rightarrow X'$ to the unique $K(g)$ such that $\pi_f \cdot K(g) = \pi_{f \cdot g}$ for all $f$. Let $\phi : H \rightarrow K \cdot H$ be the natural transformation such that $\pi_f \cdot \phi_{G_i} = \theta_f$ for all $f$.

We then "exploit" the Kan extension situation as follows: Given any $\hat{\theta} : C \rightarrow D$ such that $\Delta \hat{\theta} \cdot \gamma_C = \theta$ we can set a natural transformation $\hat{\phi} : \text{id}_{\mathfrak{C}} \rightarrow K$ given by $\pi_f \cdot \hat{\phi}_X = \hat{\theta} \cdot f$ for all $f$. Then $\phi = \hat{\phi} H$, and $\hat{\theta} = \pi_{\text{id}_C} \cdot \hat{\phi}_C$ so $\hat{\theta}$ is provided and determined uniquely by the Kan extension.

In general then, $\gamma_C$ is a colimit as required. Now ... did i really need to take those products to get this out?

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For pointwise Kan extensions the claim is almost a tautology. – Martin Brandenburg May 16 '14 at 23:13
I guess so but to compute a left Kan you need $\mathfrak{C}$ co-complete, right? – hypnocat May 16 '14 at 23:25
If you assume that, it boils down to proving a specific formula for left Kan extensions. See Proposition A.5.25 in my notes, for example. – Zhen Lin May 17 '14 at 1:31
I cant help thinking that's what the author intended after all. – hypnocat May 17 '14 at 9:25