Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

$\partial_A$ 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 $R$.

share|improve this answer
add comment

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.

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.