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Let $(\mathcal{C},\otimes, I)$ be a symmetric monoidal closed category (with $\otimes$-unit $I$).

Suppose that $\mathcal{D}$ is a (full) subcategory of $\mathcal{C}$, which contains $I$, and which is closed under $\otimes$, i.e. $A,B\in\mathcal{D}$ implies $A\otimes B\in \mathcal{D}$.

Does it follow that $(\mathcal{D},\otimes)$ becomes a symmetric monoidal closed category itself?

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  • $\begingroup$ No. What if $\mathcal{D}$ does not contain the unit of $\otimes$? Note that if we drop the closed part a special case of what you wrote is if $(M,+)$ is a commutative monoid and $S$ is a subset of $M$ which is closed under $+$, is $(S,+)$ a commutative monoid. $\endgroup$
    – Nex
    Sep 14, 2016 at 9:04
  • $\begingroup$ If we keep the closed part a special case is if $(M,+)$ is an abelian group and $S$ is a subset of $M$ which is closed under $+$ is $(S,+)$ an abelian group. $\endgroup$
    – Nex
    Sep 14, 2016 at 9:11
  • $\begingroup$ As an explicit example if we take $(\mathbb{Z},+)$ as an abelian group and $\mathbb{N}$ as a subset closed under $+$ in it. We see that $(\mathbb{Z}, +)$ is a symmetric monodical closed category (where we consider $\mathbb{Z}$ as a discrete category) but $(\mathbb{N},+)$ is not. $\endgroup$
    – Nex
    Sep 14, 2016 at 9:20
  • $\begingroup$ @Nex hey, thanks for this. Actually, even though I did not say it, I do require that the unit is in $\mathcal{D}$. Is it closed then or do we still need more assumptions? $\endgroup$
    – HugoMadderset
    Sep 14, 2016 at 9:28
  • $\begingroup$ I have edited the question. $\endgroup$
    – HugoMadderset
    Sep 14, 2016 at 9:55

1 Answer 1

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No, $\mathcal{D}$ will usually not be closed. For instance, let $\mathcal{C}=\mathtt{Set}$ and $\otimes=\times$, and let $\mathcal{D}$ be the full subcategory of countable sets. Then $\mathcal{D}$ is a monoidal subcategory, since a product of countable sets is countable. But an exponential of countable sets need not be countable, so there is no reason to expect $\mathcal{D}$ to be closed. To explicitly prove that it isn't, note that if $X$ were an internal Hom-object from $\mathbb{N}\to\mathbb{N}$, then there would need to be a map from $1$ to $X$ for each map from $1\times\mathbb{N}\to\mathbb{N}$, so $X$ would need to have uncountably many points.

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  • $\begingroup$ I see. But then, when is it? I mean, this is a very straightforward question, I would have guessed that there should be some known classification of those subcategories which inherit the monoidal closed structure... $\endgroup$
    – HugoMadderset
    Sep 14, 2016 at 10:01
  • $\begingroup$ Well, an obvious sufficient condition is that $\mathcal{D}$ is closed under forming Hom-objects. This isn't quite necessary: the necessary and sufficient condition is that for any two objects in $\mathcal{D}$, their Hom-object in $\mathcal{C}$ has a coreflection in $\mathcal{D}$. In particular, it also suffices for $\mathcal{D}$ to be a coreflective subcategory of $\mathcal{C}$. $\endgroup$
    – Eric Wofsey
    Sep 14, 2016 at 10:10
  • $\begingroup$ An instructive example where $\mathcal{D}$ is closed despite not being closed under forming Hom-objects is to take $X$ to be a topological space, $\mathcal{C}$ to be the poset $\mathcal{P}(X)$, $\otimes=\cap$, and $\mathcal{D}$ to be the poset of all open subsets of $X$. Then $\mathcal{D}$ is a coreflective subcategory, the coreflection of a subset of $X$ being its interior. The internal Hom from $A$ to $B$ in $\mathcal{C}$ is $(X\setminus A)\cup B$, and the internal Hom in $\mathcal{D}$ is then the interior of $(X\setminus A)\cup B$. $\endgroup$
    – Eric Wofsey
    Sep 14, 2016 at 10:44

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