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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!

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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.

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  • $\begingroup$ And what if we impose $\mathcal D= \mathrm{SET}$ -- the (large) category of sets? $\endgroup$ – Yon_ext Jul 21 '16 at 12:33
  • $\begingroup$ 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. $\endgroup$ – Todd Trimble♦ Jul 21 '16 at 13:28
  • $\begingroup$ 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.) $\endgroup$ – Yon_ext Jul 21 '16 at 13:45
  • $\begingroup$ 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? $\endgroup$ – Todd Trimble♦ Jul 21 '16 at 14:15
  • $\begingroup$ 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. $\endgroup$ – Yon_ext Jul 21 '16 at 14:27

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