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Let $\mathcal{C}$ be a distributive category (with finite products and finite coproducts) and let $R$ be an object of $\mathcal{C}$ such that for any object $A$ of $\mathcal{C}$ the exponential $R^A$ exists in $\mathcal{C}$. Let $\partial_A:A\to R^{R^A}$ be the morphism obtained by currying the evaluation morphism from $A\times R^A$ to $R$. Is $\partial_A$ monic? Are there some assumptions about $R$ which would make $\partial_A$ a monomorphism?

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$\partial_A$ is not always monic. Let $\mathcal C$ be the category of sets and let $R$ be a one-element set. Then $R^A$ exists and is a one-element set for all $A$, so $\partial_A$ won't be monic when $A$ has 2 or more elements.

I don't immediately see nice conditions to make $\partial_A$ monic, though in the category of sets it would suffice to require $R$ to have at least two elements. Intuitively, you want that distinct points in $A$ can be separated by maps to $R$.

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Is not your condition equivalent to saying that $R$ is a cogenerator?

[I made this edit some time ago, but as I see it has been lost in the review process (how sad!)]

No, it is not equivalent to "$R$ is a cogenerator" --- however it is clearly equivalent to "$R$ is an internal cogenerator" (the proof is along the same line as given below). Where an internal cogenerator is an object $R$ such that the functor $\hom(-, R) \colon \mathbb{C}^{op} \rightarrow \mathbb{C}$ is faithful.

The category (a Grothendieck topos) of nominal sets serves as counterexample. In this category the subobject classifier is a two-element set with a trivial action. It is easy to check that it cannot be a cogenerator. However it is an internal cogenerator.

Actually, the most interesting direction to you is easy to prove even in a more general setting --- if a monoidal category has sufficiently many linear exponents, and $R$ is a cogenerator, then the canonical $\overline{\epsilon} \colon A \rightarrow ((A \multimap R) \multimap R)$ is a monomorphism. For let us assume there are morphisms $a, b \colon X \rightarrow A$. By the definition of the exponent $((A \multimap R) \multimap R)$, $\overline{\epsilon} \circ a$ and $\overline{\epsilon} \circ b$ are in a bijective correspondence with morphisms: $$X \otimes (A \multimap R) \overset{a, b \otimes \mathit{id}}\rightarrow A \otimes (A \multimap R) \overset\epsilon\rightarrow R$$ If $a \not= b$, then by definition of a cogenerator $R$, there exists $\phi \colon A \rightarrow R$ distinguishing these morphisms, i.e. $\phi \circ a \not= \phi \circ b$. Moreover, under transposition $\phi$ corresponds to the internalized linear element $|\phi| \colon I \rightarrow (A \multimap R)$ satisfying $\epsilon \circ (\mathit{id} \otimes |\phi|) = \phi$. So if we precompose the above diagram with $$X \approx X \otimes I \overset{\mathit{id} \otimes |\phi|}\rightarrow X \otimes (A \multimap R)$$ then, we get $\phi \circ a, \phi \circ b \colon A \rightarrow R$. Since precompositions $\phi \circ a \not= \phi \circ b$, then $\epsilon \circ (a \otimes \mathit{id}) \not= \epsilon \circ (b \otimes \mathit{id})$ and by the bijective correpondence $\overline{\epsilon} \circ a \not= \overline{\epsilon} \circ b$.

The other direction should be true as well (in any closed category exponents have "enough" points), but I do not see any simple proof. Perhaps I am overlooking something obvious.

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I didn't know about cogenerators, but indeed it seems to be equivalent, and in my case it will be quite easier to prove that R is a cogenerator, compared to the monicity of the morphism d_A. Thank you. – Valentin Jan 20 '13 at 23:42
I agree with your proof, Mockup, but I was wrong, the two notion are not equivalent. Indeed if $R$ is a cogenerator then $\partial_A$ is monic, but the converse is not true ! In fact in my particular case $\partial_A$ is monic, but $R$ is not a cogenerator. I don't have time to put all the details of my category, but this is a category of games and strategies in which $R^{(R^A)}$ is much bigger than $R$ alone, there is only one morphism from $1$ to $R$, but many from $R^A$ to $R$. The idea is that a morphism from $R^A$ to $R$ contains not only the result (which is the unique inhabitant of $R$) – Valentin Jan 21 '13 at 18:02
My previous comment is not complete, here is the end : The idea is that a morphism from $R^A$ to $R$ contains not only the result (which is the unique inhabitant of $R$) as in sets, but also the computation. – Valentin Jan 21 '13 at 20:33
@Valentine, be generous and share your counterexample with us :-) I tell you, why I though these two concepts are equivalent. A cogenerator is an object $R$ such that the functor $\hom(-, R) \colon \mathbb{C}^{op} \rightarrow \mathbf{Set}$ is faithful. An internal cogenerator is an object $R$ such that the functor $R^{(-)} \colon \mathbb{C}^{op} \rightarrow \mathbb{C}$ is faithful. I though these concepts coincide. (cont) – Michal R. Przybylek Jan 22 '13 at 10:45
The first functor may be obtained by postcomposing the second one with the global section functor $\hom(1, -) \colon \mathbb{C} \rightarrow \mathbf{Set}$, i.e.: $\hom(1, -) \circ R^{(-)} = \hom(1, R^{(-)}) = \hom(-, R)$. The above proof just shows that if the composition is faithful then the precomposing functor (here: $R^{(-)}$) has to be faithful as well. – Michal R. Przybylek Jan 22 '13 at 10:46

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