Does the Yoneda Embedding Theorem really follow from the Yoneda Lemma? 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?
 A: 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 $\beta^{-1}$.
