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 $Z \to Z_i$.


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.

  • $\begingroup$ I have to say, I really like this question. It's well-written, concise, and interesting. $\endgroup$ 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
    Jan 5, 2013 at 17:50
  • $\begingroup$ For those who didn't understand Todd's insinuation: mathoverflow.net/questions/112574/… (still unsolved!) $\endgroup$ 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. $\endgroup$ Jan 7, 2013 at 8:03

1 Answer 1


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 $Z' \cap X \subseteq Y$!

  • $\begingroup$ 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. $\endgroup$ Jan 5, 2013 at 13:21
  • 2
    $\begingroup$ $Y$ 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
    Jan 5, 2013 at 15:33
  • 1
    $\begingroup$ 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
    Jan 5, 2013 at 16:26
  • 1
    $\begingroup$ ... 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
    Jan 5, 2013 at 16:31
  • 1
    $\begingroup$ Alright, now it makes sense. Thank you again! $\endgroup$ Jan 5, 2013 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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