Are Yoneda lifts (twice) of fully faithful functors fully faithful?

Question

$\newcommand{\C}{\mathcal{C}}$ $\newcommand{\D}{\mathcal{D}}$ $\newcommand{\A}{\mathcal{A}}$ $\newcommand{\S}{\mathcal{S}}$ $\newcommand{\Psh}{\mathrm{Psh}}$ $\newcommand{\Lan}{\mathrm{Lan}}$ $\newcommand{\Hom}{\mathrm{Hom}}$ $\newcommand{\Set}{\mathbf{Set}}$ $\newcommand{\yo}{\mathscr{Y}}$

If $\A$ is any category, I'll write $\yo_\A : \A \rightarrow \Psh(\A)$ the Yoneda embedding.

Let $\C$ be a small category, $\D$ be any category, and $F : \C \rightarrow \D$ be fully faithful.

Then if I define $F_1 : \Psh(\C) \rightarrow \Psh(\D)$ by the formula~:

$$ F_1 = \Lan_{\yo_\C}(\yo_\D \circ F) $$

i.e. for $A \in \Psh(\C)$~:

$$ F_1 A = \int^{c \in \C} \Hom(\yo_\C c, A) \otimes \yo_D F c = \int^{c \in \C} Ac \otimes \yo_D F c $$

and similarly $F_2 : \Psh(\Psh(\C)) \rightarrow \Psh(\Psh(\D))$ by the formula~:

$$ F_2 = \Lan_{\yo_{\Psh(\C)}\circ\yo_\C}(\yo_{\Psh(D)}\circ \yo_\D \circ F) $$

i.e. for $A \in \Psh(\Psh(\C))$~:

$$ F_2 A = \int^{c \in \C} \Hom(\yo_{\Psh(\C)} \yo_\C c, A) \otimes \yo_{\Psh(D)} \yo_D F c = \int^{c \in \C} A \yo_\C c \otimes \yo_{\Psh(D)} \yo_D F c $$

Now my question is the following : are $F_1$ and $F_2$ fully faithful ???

I seem to have a positive answer for $F_1$. Indeed :

\begin{align*} \Hom_{\Psh(\D)}(F_1 A, F_1 B) & = \int_{c\in \C} \Hom_{\Psh(\D)} \left (Ac \otimes \yo_\D Fc , F_1 B \right ) \\ & = \int_{c\in \C} \Hom_{\Set} \left (Ac, \Hom_{\Psh(\D)}(\yo_\D Fc, F_1 B ) \right) \\ & \overset{\text{Yoneda}}{=} \int_{c\in \C} \Hom_{\Set} \left (A(c), (F_1 B) F c \right) \\ & = \Hom_{\Psh(\C)}(A, F_1 B \circ F ) \end{align*}

Moreover for every $a \in \C$:

\begin{align*} (F_1 B \circ F) a & = \int^{c \in \C} Bc \otimes \yo_{\D} Fc (Fa) \\ & = \int^{c \in \C} B(c) \otimes \Hom_{\D}(Fa,Fc). \\ & = \int^{c \in \C} B(c) \otimes \Hom_{\D}(a,c) \overset{\text{Ninja Yoneda}}{=} B(a). \end{align*}

So $ F_1 B \circ F = B$, and thus~:

$$ \Hom_{\Psh(\D)}(F_1 A, F_1 B) = \Hom_{\Psh(\C)}(A, B). $$

So $F_1$ is fully faithful. If I try to do the exact same proof with $F_2$, I seem to obtain in the end~:

$$ \Hom_{\Psh(\Psh(\D))}(F_2 A, F_2 B) = \Hom_{\Psh(\C)} (A\circ \yo_\C , B \circ \yo_\C). $$

so I guess it's not fully faithful ??? But it's kinda weird, because in the case where $\Psh(\mathcal{A})$ is taken to mean $[\A:\S]$ for a certain small category $\S$, and if we have a functor $\yo'_\A : \A \rightarrow [\A:\S]$,then we certainly have (don't we ? I'm a bit confused) :

$$ \Lan_{\yo'_{\Psh(\C)}\circ \yo'_{\C}}(\yo'_{\Psh(\D)}\circ\yo'_{\D}\circ F) = \Lan_{\yo'_{\Psh(\C)}}(\yo'_{\Psh(\D)}\circ \Lan_{\yo'_{\C}}(\yo'_{\D}\circ F)) $$

and so if this operation sends embeddings on embeddings, it should still do so when applied twice !!! I'm surprised that such a thing could break just because $\Set$ is not small. I mean, everything has looked pretty formal. If $\C$ was a category where every Hom-set was smaller that some cardinal $\kappa$ (which happens), then up to equivalence you could replace $\Psh(C)$ by $[\C:\kappa]$ (which is small) and you still have a Yoneda embedding, for example... so it should be able to "stand" another extension !

TL;DR : If $F : \C \rightarrow \D$ is a fully faithful functor with $\C$ small, I seem to be able to define a fully faithful functor $F_1 : \Psh(\C)\rightarrow \Psh(\D)$ by Yoneda extension but the "next step" seems to fail and the simplest functor $F_2 : \Psh(\Psh(C)) \rightarrow \Psh(\Psh(D))$ which I can define (without taking colimits not on small categories, of course !) doesn't seem to be fully faithful : why does the pattern seem to break ?

Thanks to whoever helps me out of this ^^

(EDIT : When we look at the definition for $F_2 A$, it is clear that it only depends on $A \circ \yo$ : that's probably the problem... How should I define $F_2$ ? Why does it seem to me that when everything is small there is no problem ?)

(EDIT2 : is this a MO question ? I never feel confident to ask anything there so I don't know)

Answer 1

You $F_1$ is just the left Kan extension along $F^\text{op}$; it is going to be fully faithful for a general fact about Kan extensions: extending along a fully faithful functor gives an isomorphism $H\cong Lan_F(HF)$, and this is the unit of the adjunction $Lan_F \dashv -\circ F$, which is invertible iff the left adjoint if full and faithful.

As for your $F_2$, it does not make sense if you don't take small presheaves (the "category" $Psh(Psh(C))$ is not a "category", because it is not locally small). :-)

Answer 2

$$ \newcommand{\C}{\mathcal{C}} $$ $$ \newcommand{\D}{\mathcal{D}} $$ $$ \newcommand{\Hom}{\mathrm{Hom}} $$ $$ \newcommand{\Psh}{\mathrm{Psh}} $$

Thanks a lot to Fosco for the indication to use "small presheaves" instead of presheaves. Not only does it make what I write meaningful, but I'm also able to prove what I wanted to prove if $\mathrm{Psh}$ means small presheaves !

Take $\mathcal{C}$ not necessarily small. If $A,B$ are small presheaves on $\mathcal{C}$, then there is a small subcategory $\mathcal{C}'$ (we can take the same twice, e.g. by taking a coproduct of full subcategories) of $\mathcal{C}$ and two presheaves $A^*,B^*$ on $\mathcal{C}'$ such that~:

$$ A = \int_{c\in \C'} A^*(c) \Hom_{\C}(-,c) $$ $$ B = \int_{c\in \C'} B^*(c) \Hom_{\C}(-,c) $$

Moreover it is easy to check that $\Hom_{\Psh(\C)}(A,B) = \Hom_{\Psh(\C ')}(A^*, B^*)$ (because the inclusion of a full subcategory is fully faithful, so $A$ and $B$ are Kan extensions along a fully faithful functor...).

Now the elements of $\Psh(\mathcal{D})$ I want to define are simply~:

$$ F^* A = \int_{c\in \C'} A^*(c) \Hom_{\C}(-,Fc) $$ $$ F^* B = \int_{c\in \C'} B^*(c) \Hom_{\C}(-,Fc) $$

(it is easy to check that it is just Lan along $F^{op}$, as you noticed).

And indeed we have :

\begin{align*} \Hom(F^* A, F^* B) & = \int_{c\in \C'} \Hom(A^* c \otimes \Hom_{\C}(-,Fc), F^* B) \\ & = \int_{c\in \C'} \Hom(A^* c, (F^* B)Fc) \text{ by tensor-Hom adjunction and Yoneda lemma} \\ & = \Hom(A^*, (F^* B) \circ F) \end{align*}

and :

\begin{align*} (F^*)B F a & = \int_{c\in \C'} B^*(c) \Hom_{\C}(Fa,Fc) \\ & = \int_{c\in \C'} B^*(c) \Hom_{\C}(a,c) \text{ because } F \text{ is fully faithful}\\ & = B^* a \text{ by Ninja Yoneda} \end{align*}

So indeed~:

$$ \Hom(F^* A, F^* B) = \Hom(A^*, B^*) = \Hom(A,B) $$

so $F^*$ is fully faithful ; i.e. the Yoneda extension sends embeddings to embeddings ! And since I never supposed $\mathcal{C}$ small I can just iterate it by taking $\C_{new} = \Psh(\C)$ etc. This is exactly what I needed ! Thanks a lot !

(I guess you're gonna make a proof 2000 times simpler than mine too, but I don't think I made mistakes)