# Yoneda extension of a faithful functor is faithful

Let $F: \mathcal C \to \mathcal D$ be a functor with $\mathcal D$ cocomplete, and let $\mathscr P \mathcal C$ be the free cocompletion of $\mathcal C$ (i.e., the category of small presheaves on $\mathcal C$), so that there is a (unique, up to isomorphism) cocontinuous extension $\hat F$ of $F$ along the Yoneda embedding $\mathcal C \hookrightarrow \mathscr P \mathcal C$.

I believe that the following is true: if $F$ is faithful, so then is $\hat F$. Does anyone have a nice explanation? I'd be even happier with a proof expressible in the language of enriched category theory.

A subsidiary question. The extension $\hat F$ is the left Kan extension of $F$ along the Yoneda embedding. Is that true that any left Kan extension of a faithful functor along a fully faithful functor is again faithful? Once again, an answer in enriched category theory would be very much appreciated!

As for the first question, it's false. This is obvious in the case of posets $C, D$ (with $D$ a sup-lattice): any poset map $C \to D$ is faithful when considered as a functor, but the induced functor $Set^{C^{op}} \to D$ can't be faithful (for most presheaves $F, G$, there will be more than one natural transformation $F \to G$). Of course this answers the second question as well, since the assertion there is even stronger.
• And what if we impose $\mathcal D= \mathrm{SET}$ -- the (large) category of sets? Jul 21 '16 at 12:33
• It still doesn't work. The left Kan extension of the functor $0: 1 \to Set$, taking the unique object of $1$ to the initial object (and which is a faithful functor), is the functor $Set^1 \to Set$ that takes every object to the initial object. Jul 21 '16 at 13:28
• I think that is not a counter-example. If we identify $\mathrm{SET}^1$ with $\mathrm{SET}$ in the canonical way, then the Yoneda extension $\mathrm{SET}^1 \to \mathrm{SET}$ becomes the identity functor on $\mathrm{SET}$. (Note: your functor $\mathrm{SET}^1 \to \mathrm{SET}$ cannot be the Yoneda extension, since it is not cocontinuous.) Jul 21 '16 at 13:45
• This functor is the composite $0 \circ !: Set \to \mathbf{1} \to Set$. It has a right adjoint $1 \circ !: Set \to \mathbf{1} \to Set$ since $0 \dashv ! \dashv 1$ (where $1: \mathbf{1} \to Set$ denotes the terminal object). So $0 \circ !$, being a left adjoint, is cocontinuous. Now are you convinced? Jul 21 '16 at 14:15
• No sorry... First, $0$ is not left adjoint to $!$; $\mathrm{SET}(0\ast=1,Y) \simeq Y$ is clearly not in bijection with $\mathbf 1(\ast,!Y=\ast) = 1$ in general. Secondly, $0 \circ !$ is clearly not cocontinuous, since it doesn't respect disjoint unions. Jul 21 '16 at 14:27