Is the canonical morphism from $A$ to $R^{R^A}$ monic?

Question

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 $\partial_A \colon A\to R^{R^A}$ 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?

Answer 1

$\partial_A$ 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 $R$.

Answer 2

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.

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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.