Suppose $\mathbb{C}$ is a locally small category, and denote by $[\mathbb{C}^{\mathrm{op}}, \mathrm{Set}]$ its category of presheaves.
Define the functor $F : \mathbb{C} \to [\mathbb{C}^{\mathrm{op}}, \mathrm{Set}]$ by mapping each object $X$ to its presheaf $FX = \mathrm{Hom}(-, X)$, and each arrow $f : X \to Y$ to the natural transformation $Ff : FX \Rightarrow FY$ whose components are obtained by postcomposition with $f$, i.e. $(Ff)_Z = f \circ – : \mathrm{Hom}(Z, X) \to \mathrm{Hom}(Z, Y)$.
The Yoneda Embedding theorem states
(YET) $F$ is fully faithful.
This is often said to follow from the Yoneda Lemma, which in many places (e.g. nLab) is stated in terms of the existence of a bijection, e.g.:
(YL) There exists a bijection
$$[\mathbb{C}^{\mathrm{op}}, \mathrm{Set}](\mathrm{Hom}(-,X), \mathrm{Hom}(-,Y)) \cong \mathrm{Hom}(X, Y).$$
(I have specialised to the case of two hom functors since this is all that is needed in this question.)
In contrast, in some places (e.g. in Emily Riehl's book), I have seen an alternative statement of the Yoneda Lemma that makes explicit the form of the bijection it describes:
(YL+) The map that takes a natural transformation $\alpha : \mathrm{Hom}(-,X) \Rightarrow \mathrm{Hom}(-,Y)$ to $\alpha_X(id_X) : X \to Y$ is a bijection.
I can see how to obtain YET from YL+, since once we know that the map $\alpha \mapsto \alpha_X(\mathrm{id}_X)$ is a bijection, it is straightforward to check that this map can be inverted via $F$. (Specifically, if we denote this bijection by $\Psi$, it is straightforward to show that $\Psi(F(f)) = f$ for all $f : X \to Y$.)
However, I can't see this directly from YL, since it is not clear that the isomorphism of sets that YL refers to is in fact the one induced by $F$.
Am I missing something that would make YET a more obvious consequence of YL directly? (One thing I have left out here is the naturality of the bijections in YL and YL+ – would this help?) Or is YL+ really what is needed to obtain YET without effectively reproving the Yoneda Lemma from scratch?
Best Answer
There is an interesting variation of the Yoneda lemma at play here.
Note that a transformation with components $\phi_{X,Y}\colon C(X,Y)\to D(FX,FY)$ given by $\phi_{X,Y}:f\mapsto Ff$ is natural if and only if $F$ is a functor. Moreover, $F$ is full, resp. faithful, resp. fully faithful, if and only if the components are surjections, resp. injections, resp. bijections
The variation is then this: any natural transformation $\alpha\colon C(X,Y)\to D(FX,FY)$ is of the form $D(\beta,FY)\circ\phi=D(FX,\beta)\circ\phi$ for a natural transformation $\beta$ from $F$ to itself with components $\beta_X=\alpha_{X,X}(\mathrm{id}_X)$.
Indeed, naturality in $Y$ of $\alpha\colon C(X,Y)\to D(FX,FY)$ implies $\alpha_{X,Y}(f)=\alpha_{X,Y}\circ C(X,f)(\mathrm{id_X})=D(FX,Ff)\circ\alpha_{X,X}(\mathrm{id}_X)=Ff\circ\beta_X$ where $\beta_X=\alpha_{X,X}(\mathrm{id_X})\in D(FX,FY)$, while naturaliry in $X$ implies $\alpha_{X,Y}(f)=\beta_Y\circ Ff$. In particular, $Ff\circ\beta_X=\alpha_{X,Y}(f)=\beta_Y\circ Ff$, asserts exactly that $\beta$ is a natural transformation from $F$ to itself such that $D(\beta,FY)\circ\phi=\alpha=D(FX,\beta)\circ\phi$.
We now claim that if $\alpha_{X,X}$ are surjective, then $\beta$ is a natural isomorphism from $F$ to itself, whence $D(FX,\beta)$ and $D(\beta,FY)$ are natural bijections from $D(FX,FY)$ to $D(FX,FY)$ by which $\alpha$ and $\phi$ are related. In particular, $\alpha$ a natural bijection implies $\phi$ is a natural bijection, e.g. (YL-wtih-naturality) implies (YET).
Indeed, if $\alpha_{X,X}$ are surjective, then $\mathrm{id}_{FX}=Fs\circ j_X=j_X\circ Ft$ for some $s,t\in C(X,X)$, whence $Fs=Ft=\beta_X^{-1}$ are the unique two-sided inverses of $\beta_X$, from which follows that the natural transformation $\beta$ has an inverse $\beta^{-1}$.