Codense if and only if truncated Yoneda embedding is fully faithful

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.

• Continuous or fully faithful? – Fosco Jun 28 '16 at 15:54
• @FoscoLoregian sorry, fixed. – Arrow Jun 28 '16 at 19:31
• I answered you (-: – Fosco Jun 29 '16 at 8:32

1 Answer

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!

• Can 1.29 be proved without mentioning wedges? – Arrow Jun 29 '16 at 12:14
• Sure, $Nat(H,K)$ is simply the equalizer of the pair of arrows $$\prod_{c\in C}\hom(Hc,Kc) \rightrightarrows \prod_{f\colon c\to c'}\hom(Hc, Kc')$$ where the pair is induced by pre- or post-composition with an arrow $Hf, Kf$. This is indeed only an equivalent way to characterize the universal property of the end $\int_c\hom(Hc,Kc)$. :) – Fosco Jun 29 '16 at 16:22