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Suppose $\mathcal{C}$ is a Cartesian Closed Category. Letting $B^A$ denote the internal hom of $A$ and $B$ and $ev_B^A:B^A\times{}A\to{}B$ denote the evaluation morphism, there is a bifunctor

\begin{align*} E:\mathcal{C}^{op}\times\mathcal{C}&\to\mathcal{C}\\\\ (A,B)&\mapsto{}B^A\\\\ (f:A\stackrel{\mathcal{C}^{op}}{\to}{}A',g:B\to{}B')&\mapsto{}g^f:B^A\to{}(B')^{(A')} \end{align*}

where $g^f$ is the unique morphism such that


This functor is analogous to the hom bifunctor in many respects. In particular, we may transpose it to get the analogues to the covariant and contravariant Yoneda embeddings

\begin{align*} \mathcal{C}&\to{}\mathcal{C}^{\mathcal{C}^{op}}\\\\ B&\mapsto{}B^{(-)}\\\\ (g:B\to{}B')&\mapsto{}(g^A:B^A\to{}(B')^A)_{A\in{}Ob(\mathcal{C})} \end{align*}


\begin{align*} \mathcal{C}^{op}&\to{}\mathcal{C}^{\mathcal{C}}\\\\ A&\mapsto{}(-)^A\\\\ (f:A\stackrel{\mathcal{C}^{op}}{\to}{}A')&\mapsto{}(B^f:B^A\to{}B^{A'})_{B\in{}Ob(\mathcal{C}^{op})} \end{align*}


1) Neither of these functors are injective on objects. Consider the complete preorder on a nonempty set $X$ as a category. Now take a single selected element $x_0\in{}X$ as the product and exponential of every pair of objects.

2) The usual methods can be used to show that that each of these functors is faithful.

My question is the following: under what conditions can we conclude these functors are full?

One can attempt to use the usual argument, but I have only been able to show that the (necessarily unique) preimage of a natural transformation recovers the original transformation on $1$-elements (here, $1$ denotes a terminal object). This is because $B^f$ does not act exactly as precomposition, except on $1$-elements. As a result, these functors will be full if $1$ is a separator in $\mathcal{C}$, but this is a strong condition and I would be interested in other criteria.

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Surely, we should be thinking of $\mathcal{C}^{\mathcal{C}^\textrm{op}}$ as a $\mathcal{C}$-enriched category here? – Zhen Lin Aug 29 '12 at 2:28
Zhen, $\mathbb{C}^{\mathbb{C}^{op}}$ may not be $\mathbb{C}$-enriched (due to size issues and completeness of $\mathbb{C}$). – Michal R. Przybylek Aug 29 '12 at 11:23
However, you are right that the usual category $\mathbb{C}^{\mathbb{C}^{op}}$ is not the most natural choice here (if we took $\mathbb{C}^{\mathbb{C}^{op}}$ to be the category of $\mathbb{C}$-enriched functors and $\mathbb{C}$-enriched natural transformations then the internal hom functor would give a full and faithful embedding by the weak Yoneda lemma; moreover, the "internal Yoneda" from the question factors through the "weak" one; therefore the real deal is how faithful the underlying functor $\hom(1, -)$ is). – Michal R. Przybylek Aug 29 '12 at 12:52

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