# Comma category construction?

I am getting closer to the essence of a comma category.

Given 1) category $$\mathcal{A}$$ with objects $$A, B$$ and signle not-identity morphism $$f: A \mapsto B$$, and 2) single-object category with single (identity) morphism $$\mathcal{1}$$, I am trying to construct $$(F / G$$), where $$F$$ is an identity functor on the $$\mathcal{A}$$ and $$G: \mathcal{1} \mapsto \mathcal{A}$$ always selects, say, the $$A \in Obj(\mathcal{A})$$.

So there exist two objects: $$(A, \circ \in Obj(\mathcal{1}), F(A) = A \mapsto G(\circ) = A)$$ and $$(B, \circ \in Obj(\mathcal{1}), F(B) = B \mapsto G(\circ) = A)$$. Here it feels wrong to me: $$F(B) \mapsto G(\circ) = B \mapsto A$$ must be some morphism in the $$\mathcal{A}$$ according to the defintion; however such morphism does not exist.

So, I got two simple questions here: 1) What did I do wrong above? 2) What does comma category say to us? How do we "read" it, what kind of information it packs about participating functors and underlaying categories?

## 1 Answer

Given functors $$F:\mathcal{C}\to\mathcal{D}$$ and $$A:1\to\mathcal{D}$$, the objects of $$(F\downarrow A)$$ are pairs $$(X,f)$$ with $$X\in\mathrm{obj}(\mathcal{C})$$ and $$f\in\mathcal{D}(F(X),A)$$. If there aren't any morphisms $$F(X)\to A$$, then there is no such pair, and thus no such object.

• In the above example, the only object in $$(id_\mathcal{A}\downarrow A)$$ is $$(A,id_A)$$; it essentially the trivial category.

• In fact if you took the natural numbers and their ordering as a category, and let $$0:1\to\mathbb{N}$$ be the functor that picks out $$0$$, then $$(id_\mathbb{N}\downarrow 0)$$ would also have only one object and the identity arrow; infinitely many objects of $$\mathbb{N}$$ would play no role whatsoever in the comma category.

• If we considered $$\mathbb{N}$$ as a discrete category (where the only morphisms are identities) and let $$o:\mathrm{Odd}(\mathbb{N})\to\mathbb{N}$$ be the inclusion of the odd numbers, then the comma category $$(o\downarrow 0)$$ has no objects at all--it is the empty category.

Point being that $$\mathcal{C}$$ can have all kinds of objects, without there being any guarantee that the comma category is non-empty; much less that every object of $$\mathcal{C}$$ has some associated object the comma category.