Density of Yoneda Embedding - strange conclusion? I'm trying to prove $y:\mathsf C\longrightarrow \widehat{\mathsf C}$ is dense. For this it suffices to prove the composite
$$\widehat{\mathsf C}\overset{\hat y}{\longrightarrow}\widehat{\widehat{\mathsf C}}\overset{y^\ast}\longrightarrow \widehat{\mathsf C}$$ is fully faithful, where the latter is precomposition with $y^\text{op}$.
The (extension of the) Yoneda lemma says the isomorphism $\widehat{\mathsf C}(y_c,G)\cong G(c)$ is natural in $c,G$, which means there's an isomorphism of functors
$$\widehat{\mathsf C}(y^\text{op}_{(-)},-)\cong \operatorname{eval}:\mathsf C^\text{op}\times \widehat{\mathsf C}\longrightarrow \mathsf{Set}.$$
Now evaluation is the transpose of the identity while the former functor is the transpose of $y^\ast\circ \hat y$, so I seem to get that $y^\ast\circ \hat y\cong 1$, which is way more than saying it's fully faithful. What is my mistake and can my proof still work?
 A: There is no mistake. Yet another equivalent formulation of the definition of a dense functor $F : \mathsf{C} \to \mathsf{D}$ is that the restricted Yoneda embedding $\mathsf{D} \to \hat{\mathsf{C}} = [\mathsf{C}^{\mathrm{op}}, \mathsf{Set}]$, given by $$d \mapsto (c \mapsto \hom_\mathsf{D}(F(c),d)),$$ has to be fully faithful. But in your case $\mathsf{D}$ itself is $\hat{\mathsf{C}}$, and the restricted Yoneda embedding maps $F \in \hat{\mathsf{C}}$ to
$$c \mapsto \hom_{\hat{\mathsf{C}}}(y(c), F) \cong F(c),$$
i.e. it maps $F$ to $F$. In other words, it's the identity.
A: $\let\mathcal\mathsf$
Let me call an arity a functor $i\colon \mathcal A \to \mathcal E$ which is both fully faithful and dense. Remark that $y$ being already fully faithful, it is an arity if and only if it is dense.
I like the following "point of view" on arities:

Lemma. A functor $i\colon \mathcal A \to \mathcal E$ is an arity if and only if there is a fully faithful $N \colon  \mathcal E \to \hat{\mathcal A}$ such that $Ni$ is the Yoneda embedding.

The proof is quite easy: if it is an arity, choose $N$ to be the restricted Yoneda embedding $\mathcal E \to \hat{\mathcal A}$ ; it is fully faithful by hypothesis and for any $a\in \mathcal A$,
$$ Ni(a) = \mathcal E(i(-),i(a)) \simeq A(-,a)= y_a $$
Conversely, if such an $N$ exists, then first $i$ is fully faithful because $N$ and $Ni$ are, and next for $e \in \mathcal E$,
$$ N(e) \simeq \hat{\mathcal A}(y_{-},N(e)) \simeq \hat{\mathcal A}(Ni(-),N(e)) \simeq \mathcal E(i(-),e) $$
meaning that $N$ is the restricted Yoneda embedding, hence showing that $i$ is dense.
Hence morally, $\mathcal A \hookrightarrow \mathcal E$ is an arity if $\mathcal E$ is a full subcategory of $\hat{\mathcal A}$ containing all the respresentables. Then $y$ is readily an arity (hence dense). So is $\mathrm{id}_{\mathcal C}$! In some sense, those are the extremal arities. Of course, there could be many more in between (e.g. $\mathsf{Bij} \to \mathsf{Set}$ and $\Delta \to \mathsf{Cat}$ are arities).
(I don't claim this is any different in the calculus than Najib Idrissi's answer. I however hope this presentation makes you understand what's going on.)

Of course, I'm not making the terminology up here. For a compendium article on the subject, you can read Monads with arities and their associated theories.
