# Structure of a poset of subcategories

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?

• 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". Aug 12, 2019 at 23:09
• 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. Aug 13, 2019 at 8:41

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 $$h$$.