Codense if and only if truncated Yoneda embedding is fully faithful

Question

I want to prove a functor $F:\mathsf C\rightarrow \mathsf D$ is codense if and only if the truncated (dual of the) Yoneda embedding $$\mathsf D^\text{op} \rightarrow [\mathsf D,\mathsf{Set}] \to [\mathsf C,\mathsf{Set}]$$ is fully faithful. (The right functor is composition with $F$.)

By definition, a functor $F$ is codense if every object $d$ is the colimit of $F\circ \varphi_d$, where $\varphi_d:\int \mathsf D(d,-)\cong d/F \rightarrow \mathsf C$ is the discrete opfibration associated with $\mathsf D(d,-)$.

I can see that by density $h_d\cong \varprojlim\nolimits ^{\mathsf D(d,-)}\!F^\ast (h_d)$, but I don't see how to use this to show the truncated embedding is fully faithful. Since it looks like we're looking for an inverse to precomposition, I'd also appreciate if someone pointed out the exact relationship to Kan extensions.

Answer 1

By definition, a functor $G : C\to D$ is full and faithful iff $C(c,c')\cong D(Gc,Gc')$.

Again by definition, a functor is codense iff $\text{Id}_D\cong \text{Ran}_FF$, i.e. iff $d\cong \int_c Fc^{D(d,Fc)}$ (coends, p. 16), naturally in $d\in D$.

Let's put these two definitions together to prove the result: the only additional assumption I need is that $D$ admits arbitrary products.

$$ \begin{align} \text{Nat}(F^*{\bf y}(x), F^*{\bf y}(y)) &\cong \text{Nat}(D(x,F-),D(y,F-))\\ (1)&\cong \int_c {\bf Set}(D(x,Fc),D(y,Fc))\\ (2)&\cong \int_c D(y, Fc^{D(x,Fc)})\\ (3)&\cong D(y, \int_c Fc^{D(x,Fc)})\\ (4)&\cong D(y, \text{Ran}_FF(x))\\ (5)&\cong D(y,x) \end{align} $$ where

1. is coends, 1.29;
2. This is where you need arbitrary products to exist in $D$, since $Fc^{D(x, Fc)} \cong \prod_{f\in D(x, Fc)} Fc$;
3. is coends, 1.27
4. is coends, p. 16 (the description of a right Kan extension as an end)
5. follows from the assumption that $Ran_FF\cong Id$ (see also the $n$Lab page about codense functor, or the chapter dedicated to codensity monads in Dubuc's "Kan extensions in enriched category theory".

Cheers!