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Background

By a cocomplete symmetric monoidal category $C$ I mean a symmetric monoidal category whose underlying category is cocomplete and such that $- \otimes X : C \to C$ is cocontinuous for all $X \in C$. Recall that the internal hom $\underline{\mathrm{hom}}(X,-)$ is defined, if it exists, as a right adjoint of $- \otimes X$, and that $C$ is called closed if internal homs exist for all $X \in C$. According to the General Adjoint Functor Theorem, this reduces just to a size condition: For all $X,Y \in C$ there should be a set of morphisms $(Z_i \otimes X \to Y)_{i \in I}$, such that any other morphism $Z \otimes X \to Y$ factors as $Z \otimes X \to Z_i \otimes X \to Y$ for some $i \in I$ and $Z \to Z_i$.

Questions

Q1. What is a nice example of a cocomplete symmetric monoidal category which is not closed?

A standard example for a cartesian category which is not closed is $\mathsf{Top}$; but $-\times \mathbb{Q}$ doesn't preserve coequalizers so that this doesn't answer the question (similar problems with other standard examples). If Q1 is too easy, what about cartesian categories?

Q2. What is a nice example of a cocomplete category with products, such that $-\times X$ is cocontinuous for all $X$, but has no right adjoint in general?

A weaker question would be:

Q3. What is a nice example of a cocomplete symmetric monoidal category whose underlying category is not locally presentable?

Again the standard examples of non-locally presentable categories which I have found in the literature don't fit here.

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    $\begingroup$ I have to say, I really like this question. It's well-written, concise, and interesting. $\endgroup$
    – David White
    Commented Jan 5, 2013 at 17:30
  • $\begingroup$ Yes, it's a good model for one way of writing good MO questions. Now, if only we could find a cocomplete but not complete abelian category. :-) $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 17:50
  • $\begingroup$ For those who didn't understand Todd's insinuation: mathoverflow.net/questions/112574/… (still unsolved!) $\endgroup$
    – Martin Brandenburg
    Commented Jan 5, 2013 at 18:32
  • $\begingroup$ Re: Q3, there are plenty of examples in topology of even closed bicomplete symmetric monoidal categories that are not locally presentable. For instance, the category of compactly generated Hausdorff spaces. $\endgroup$
    – Mike Shulman
    Commented Jan 7, 2013 at 8:03

1 Answer 1

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Here is an amusing example which addresses Q2: take the universe $V$ of sets in a model of ZFC, as a class partially ordered by inclusion of sets. Consider a partially ordered class to be a category in the usual way. Of course, by Cantor's theorem, there is no terminal object in this category, but anyway we can freely adjoin one; let $V_+$ denote the result. Notice that cartesian products are given by taking intersections.

The category $V_+$ is small cocomplete since we can take small unions, and of course intersections $- \cap X$ distribute over unions, so we get a cocomplete cartesian monoidal category. But I claim that unless $X$ is the top element (that we freely adjoined), there is no exponential $Y^X$ for any strict subset $Y \subset X$. Indeed, if $Z = Y^X$, then $Z$ would have to be the largest set such that $Z \cap X \subseteq Y$. But there is no such largest set, since to any such set we could add more elements which do not belong to $X$ to get a larger set $Z'$, and we'd still have $Z' \cap X \subseteq Y$!

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  • $\begingroup$ Thank you for this example, which is in fact quite amusing and shows how pathological non-accessible categories can be. Two questions: A) We don't need $Y \subset X$ for your argument, right? $Y$ could be any set, even $X$ or empty. B) Isn't there a problem with the construction of $V_+$? Namely, by a $V$-category $C$, I mean a subset $\mathrm{Mor}(C)$ of $V$ together with certain operations and properties. But why should $\mathrm{Mor}(V_+)$ be a subset of $V$? If this is not the case, we could choose a bigger universe $V'$ and consider $V^+$ as a $V'$-category, but it won't be cocomplete. $\endgroup$
    – Martin Brandenburg
    Commented Jan 5, 2013 at 13:21
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    $\begingroup$ $Y$ can't be any set; if $Y$ contains $X$ then we can show $Y^X$ is the top element $1$ (since clearly $1 \cap X \subseteq Y$, and $1$ is the maximal element). But you're right that we don't need $Y \subset X$ precisely; I think all we need is that $X$ contains some elements not belonging to $Y$. As for question (B), what's the problem? I'm simply defining a category whose objects are elements of $V$ plus an extra object $1$; we have a (unique) morphism $x \to y$ if either $x \subseteq y$ in $V$ or if $y = 1$. (It's easy enough to code up all the data in $V$, if you insist on this.) $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 15:33
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    $\begingroup$ Well, one just works with isomorphs instead, using various ugly hacks. So instead of defining objects to be elements $x$ of $V$, define them to be e.g. ordered pairs $(\emptyset, x)$ where $x$ ranges over elements of $V$, and define $1$ to be something dumb like {$\emptyset$}. And define morphisms to be elements of the disjoint union of two classes where the first class consists of ordered triples $(1, x, y)$ such that $x \subseteq y$, and where the second class consists of, I don't know, elements of $(2, x)$ (which are supposed to stand for arrows $x \leq 1$). Under some such kludgy coding... $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 16:26
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    $\begingroup$ ... you can define objects and morphisms of a category isomorphic to the $V_+$ as I described it as elements of $V$, and define rules for composition, etc., etc. Just think of this as defining a partial order $V'$ in some way so that $V'$ has a terminal object $\top$ and the sub-partial order given by the complement of $\top$ is isomorphic to $V$ under the subset inclusion relation; you don't have to think of the partial order on $V'$ as literal subset inclusion itself (which is where I think the confusion arose). $\endgroup$
    – Todd Trimble
    Commented Jan 5, 2013 at 16:31
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    $\begingroup$ Alright, now it makes sense. Thank you again! $\endgroup$
    – Martin Brandenburg
    Commented Jan 5, 2013 at 16:38

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