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Say $\mathscr{A}$ is a reflective subcategory of $\mathscr{B}$, meaning the inclusion functor $i: \mathscr{A} \to \mathscr{B}$ is fully faithful and admits a left adjoint, and $\mathscr{B}$ is enriched over itself, where for sake of simplicity we assume the tensor bifunctor to be the product.

Moreover, assume $\mathscr{A}$ is closed under this tensor, and $i$ commutes with it (again, the case of products suffice).

What can we say about a self-enrichment of $\mathscr{A}$?

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    $\begingroup$ Perhaps you are looking for the concept of an exponential ideal. $\endgroup$
    – Zhen Lin
    Commented Mar 29, 2016 at 13:45

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The most natural question to ask is whether $\mathcal{A}$ is closed under the internal-hom of $\mathcal{B}$, i.e. that $[A,B]\in \mathcal{A}$ whenever $A,B\in\mathcal{A}$. This is an extra assumption, but it's sometimes easier to verify by reformulating it in terms of the tensor product; the resulting condition is that the reflection is a Hopf adjunction.

A better-known but stronger condition than this is that $[A,B]\in \mathcal{A}$ whenever $B\in\mathcal{A}$ (but $A\in \mathcal{B}$ is arbitrary). As Zhen mentioned, in the cartesian case at least this is called being an exponential ideal, and is equivalent to asking that the reflector be strong monoidal (preserve products).

It is, of course, technically possible that $\mathcal{A}$ is closed but that its internal-hom is not induced from $\mathcal{B}$, but it's hard to say anything in generality about when that might happen.

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  • $\begingroup$ If I understand your answer correctly you are saying that the closure of $\mathscr{A}$ under the internal-hom of $\mathscr{B} $ follows by the reflection being a Hopf adjunction, which implies (by def., on nlab) that the left adjoint is strong monoidal. So how is the second part stronger than the first one? In any case, thanks for your answer, it is the kind of thing I was looking for :) $\endgroup$
    – Edoardo Lanari
    Commented Mar 30, 2016 at 0:06
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    $\begingroup$ If the reflection $F: \mathcal{B} \to \mathcal{A}$ is a Hopf adjunction it only implies that the colax structure map $u_{X,Y}:F(X \otimes Y) \to F(X) \otimes F(Y)$ is an isomorphism when $Y$ belongs $\mathcal{A}$, not in general. This condition is enough to make the inner hom $[Y,Z]$ belong to $\mathcal{A}$ whenever $Y,Z$ both belong to $\mathcal{A}$. On the other hand, if $u_{X,Y}$ is always an isomorphism then we only need $Z$ to be in $\mathcal{A}$ and we get the stronger condition. $\endgroup$
    – Yonatan Harpaz
    Commented Mar 30, 2016 at 19:25

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