Does the functor $X\times-:\mathbf{Loc}\to\mathbf{Loc}$ preserve small colimits for all locales $X$?

The reason that I'm interested in this question is that the same property fails in the category of topological spaces. If products preserve colimits in a total category then it must be cartesian closed. The category $\mathbf{Top}$ is total, but fails to be cartesian closed because products don't preserve colimits. So it would be curious if $\mathbf{Loc}$ failed to be cartesian closed for the complementary reason. It's not total, so maybe products do preserve colimits in $\mathbf{Loc}$?

Another reason to ask this question is that the definition of "locale" requires precisely that products preserve colimits in its frame of opens. So it would be interesting if this property was mirrored at the category level.


1 Answer 1


I think so? But I'm putting together results I don't understand in much detail. Let's try the following:

Equivalently the question is whether taking coproduct with a frame $X$ preserves limits of frames. We'll write the coproduct of frames as $\otimes$, since it turns out to coincide with the tensor product of suplattices (see this nLab page and this nLab page). The tensor product of suplattices has right adjoint given by the hom suplattice, so preserves colimits of suplattices in both variables; I think this implies that it preserves limits in both variables, because I think limits of suplattices can be computed as colimits and vice versa (e.g. products are coproducts); this is the step I'm least confident in. If that's true, then because limits of frames are computed as limits of suplattices (which in turn are computed as limits of underlying sets), tensor product preserves limits.

At the very least I'm confident that tensor product preserves products, so it remains to check if it preserves equalizers, which maybe can be done by hand.

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    $\begingroup$ Could it be that $(\mathbf{Sup},\otimes)$ is compact closed? Then it would be obvious that $X\otimes -$ preserves limits, since it would be left and right adjoint to $X^*\otimes -$. I've found some sources that refer to "dualizable suplattices" which suggests that they think there exist some nondualizable suplattices, but I can't find an example of such. $\endgroup$ Sep 18, 2019 at 6:39
  • $\begingroup$ One point in favour of the idea that $\mathbf{Sup}$ is compact closed is the fact that it's the Eilenberg-Moore category of the monad for which the Kleisli category is $\mathbf{Rel}$, a classic example of a compact closed category. Even if $\mathbf{Sup}$ isn't compact closed we might be able to use the relationship with $\mathbf{Rel}$ to prove that $X\otimes -$ preserves limits. $\endgroup$ Sep 18, 2019 at 6:42
  • $\begingroup$ Separately, how do you know that limits of frames are computed as limits of suplattices? Is there a left adjoint to $U:\mathbf{Frame}\to\mathbf{Sup}$? $\endgroup$ Sep 18, 2019 at 6:44
  • $\begingroup$ @Oscar: oof, I'm not at all confident of the shaky step above anymore, but I'm a bit more confident that limits of frames are computed as limits of suplattices, because they're both computed as limits of sets, because they're both monadic over $\text{Set}$. But I do in addition believe that the forgetful functor from frames to suplattices has a left adjoint, and moreover I believe that it's monadic, basically because I believe frames are the models of a particular algebraic theory over suplattices (idempotent commutative monoids or something). $\endgroup$ Sep 18, 2019 at 7:07
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    $\begingroup$ $\mathbf{Sup}$ isn't compact closed. The double-dual of $\{\{a,b,c\},\{a\},\{b\},\{c\},\{\}\}$ is $\{\{a,b,c,d,e\},\{a,b,c,d\},\{a,d\},\{b,d\},\{c,d\},\{d\},\{\}\}$. $\endgroup$ May 14, 2020 at 15:40

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