# Name for a certain type of derived category similar to $(\mathbf{C}\downarrow\mathbf{C})$

$$\newcommand{Hom}{\operatorname{Hom}}$$

Consider a category $$\mathbf{C}$$, and draw a graph according to the following rules:

1. Place all objects onto the graph
2. Between every two objects $$A,B\in\operatorname{obj}(\mathbf{C})$$, if $$\Hom(A,B)\neq\emptyset$$, draw an arrow between $$A, B$$. Take that arrow to represent the entire hom-set of morphisms between $$A$$ and $$B$$.
3. Now take the line-graph of the above graph and use it to define a category $$\mathbf{C}^\prime$$, whose objects are collections of morphisms in $$\mathbf{C}$$.

Perhaps more clearly, suppose that commutative squares like the following occur in $$\mathbf{C}$$:

$$\require{AMScd} \begin{CD} A @>{f}>> A^\prime\\ @VV{h}V @VV{h^\prime}V \\ B @>{f^\prime}>> B^\prime \end{CD}$$

Then we take the entire hom-sets $$\Hom(A,B), \Hom(A^\prime,B^\prime)$$ to be objects $$H, H^\prime$$ respectively in $$\mathbf{C^\prime}$$. Furthermore, $$F=\langle f, f^\prime\rangle$$ is a morphism in $$\mathbf{C^\prime}$$; in general, a morphism $$\langle f, f^\prime\rangle$$ exists in $$\Hom(H,H^\prime)$$ iff there exists a pair of arrows $$h\in H, h^\prime\in H^\prime$$ so as to make the above diagram commute. In that regard, $$F\circ G$$ is meaningful iff the targets of $$f,f^\prime$$ are respectively the domains of $$g,g^\prime$$; in other words if both pairs share an object each. In that sense the objects of $$\mathbf{C}$$ define the morphisms of $$\mathbf{C^\prime}$$, thus the identification with the line graph of the diagram of $$\mathbf{C}$$.

In particular and for example, if $$\mathbf{C}$$ is preadditive, then all objects in $$\mathbf{C^\prime}$$ are additive abelian groups, and the morphisms $$\mathbf{C^\prime}$$ are all group homomorphisms.

Now, it's entirely possible I'm misunderstanding what comma categories are; but as I understand it, the category $$(\mathbf{C}\downarrow\mathbf{C})$$ has as its objects individual morphisms of $$\mathbf{C}$$, which isn't quite what I'm looking for. Am I wrong? If not, is there an existing name for categories as I've described them above?

• I don't understand your step 3 at all. What are the objects and morphisms of $C'$? You say "collections of morphisms" - which collections? Commented Jan 5, 2021 at 5:21
• @QiaochuYuan Actually, probably a simpler way to state it would be: the objects of $\mathbf{C}^\prime$ are the hom-sets of $\mathbf{C}$. Commented Jan 5, 2021 at 5:32
• Okay. What are the morphisms? Commented Jan 5, 2021 at 5:49
• I guess step 2 is a change of base functor from a Set-category to a Bool-category. (I haven't checked that works.) Then step 3 would just be forming the arrow category for the resulting Bool-category. Commented Jan 5, 2021 at 8:49
• Hmm... $$\require{AMScd} \begin{CD} A @>{f}>> A^\prime @>{f^\prime}>> A^{\prime\prime}\\ @VV{h}V @V{h^\prime_1}V{h^\prime_2}V @VV{h^{\prime\prime}}V \\ B @>>{g}> B^\prime @>>{g^\prime}> B^{\prime\prime} \end{CD}$$ Okay, I see your point. I think I need to require that the category has all equalizers for this to make sense -- then I think I can conclude that the larger square does in fact commute? Commented Jan 7, 2021 at 12:05