8
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 limits of frames are computed as limits of suplattices (which in turn are computed as limits of underlying sets), so 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.

$\endgroup$

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.