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?
Best 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.