1
$\begingroup$

Given a category $\mathbf{C}$, we can consider monomorphisms into it. These are the faithful and injective-on-objects functors (this violates the principle of equivalence). The idea is to try to get a lattice of subobjects.

Pullbacks (in the 1-categorical sense, which also has problems with principle of equivalence) preserve monomorphisms, so we can have a notion of "meet". Then, given monomorphisms $\mathbf{X} \to \mathbf{C}$ and $\mathbf{Y} \to \mathbf{C}$, we can take the image of the coproduct $\mathbf{X} + \mathbf{Y} \to \mathbf{C}$ and have a notion of "join". Here the image is the smallest subcategory containing all the images of morphisms under the functor (again, not really treating $\mathbf{Cat}$ as a 2-category).

Is this construction plausible or is there any problem I am overlooking? Does that poset have any extra structure? I am not sure if I should understand from here that for it to be a distributive lattice we would need $\mathbf{Cat}$ to be coherent, but it is not even regular. If not, can we impose some conditions on $\mathbf{C}$ to get some structure?

$\endgroup$
2
  • 1
    $\begingroup$ Yes, this is the general construction in any complete category of the lattice of subobjects of a fixed object, if we interpret "image" as "least subobject through which a morphism factors". $\endgroup$ Aug 12, 2019 at 23:09
  • $\begingroup$ Thanks! then the only thing I do not see now is why nlab asks for the category to be coherent to have a lattice of subobjects. $\mathbf{Cat}$ is not coherent but that construction seems to work anyway. $\endgroup$ Aug 13, 2019 at 8:41

1 Answer 1

3
$\begingroup$

Like Kevin said in the comments, your definition of join should work just fine in any category with finite coproducts and a good notion of "image." The issue with $\textbf{Cat}$ is that you won't have a distributive lattice (which is specifically what the nLab says a coherent category will give you).

Here's an example of distributivity failing: Let $\mathcal{C}$ be a category with three objects $x,y,z$ and three non-identity morphisms, $f\colon x\to y$, $g\colon y\to z$, and $h=g\circ f$. By abuse of notation, I'll refer to subcategories by the objects or morphisms which generate them. If we take the join of $f$ and $g$ we get all of $\mathcal{C}$, and so taking the meet of $h$ with that we get $h$. But, if we instead take the meet of $h$ with $f$ and $g$ individually first, we get $x$ and $z$, and then the join of $x$ and $z$ is $x\cup z$, which does not include $h$.

$\endgroup$

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.