As far as I understand, cone is a pair of some object $X \in Obj(\mathcal{C})$ (which can be viewed as a $\Delta_X$ - constant functor from some other category to the $\mathcal{C}$) and a set of morphisms in $\mathcal{C}$ obeying a known requirement, which I ignore here.

So, by the definition cone is $(X \in Obj(\mathcal{C}), M \subset Ars(\mathcal{C}))$ (once again, commutation requirement ignored).

Now my textbook says that comma category $(\Delta | F)$, where $F$ is a functor mapping somewhat category $\mathcal{C'}$ to the $\mathcal{C}$, produces a category of cones. More than that, initial object in a $(\Delta | F)$ is a limit to the $F$. I can't make it out:

Objects of comma category are triplets, each ships single morphism rather than a set of those. But cone requires a set of morphisms. Therefore, to gather a cone (by given definition) one has to select a subcategory of $(\Delta|F)$ and "merge" each morphism of the selected triplet into a resulting set of morphisms.

Thus, I am confused with what is called a "cone" in the category theory? And how does it relate to the category of cones (special case of comma category)?

  • $\begingroup$ What textbook is it? $\endgroup$
    – Arnaud D.
    Nov 12 '18 at 9:24

Writing $(\Delta \mid F)$ is a slight abuse of notation. Remember that $\Delta$ is a functor $\mathcal C \to \mathcal C^{\mathcal C'}$, while $F$ is a functor $\mathcal C' \to \mathcal C$.

To form a comma category $(G \mid H)$, the functors $G$ and $H$ need to have the same codomain.

So, we actually consider $F$ as a functor $\bf1 \to \mathcal C ^{\mathcal C'}$. Let's write it as $\hat F :\bf 1 \to \mathcal C ^{\mathcal C'}$, where $\hat F \star = F$ (and $\star$ is the unique object of $\bf 1$).

Now we see that the objects of $(\Delta \mid \hat F)$ are triples $(A, \star, h)$, where $A$ is an object of $\mathcal C$ and $h$ is a morphism $\Delta A \to \hat F\star = F$. So $h$ is a natural transformation between the functors $\Delta A$ and $F$.

It is not hard to verify that this is the same thing as a cone from $A$ to $F$.

Edit: By “cone to $F$”, I mean an object $A$ of $\mathcal C$, and a family of morphisms $\gamma_X :A \to F X$, indexed by objects $X$ of $\mathcal C’$, such that for every $f : X \to Y$, we have $\gamma_Y = Ff \circ \gamma_X$.

A morphism of cones $(A, \{\gamma_X\}_X) \to (B, \{\delta_X\}_X)$ is a morphism $f : A \to B$ such that for every X we have $\gamma_X = \delta_Y \circ f$.

So given a category $\mathcal C$ and a functor $F : \mathcal C’ \to \mathcal C$ we can form the category of cones to $F$ in $\mathcal C$. This is isomorphic to the comma category $(\Delta \mid \hat F)$.

  • $\begingroup$ Kindly asking to update your post with a final, rigorous definition of what "cone" is. In particular - does each triple $(A, \star, h)$ define a cone? $\endgroup$
    – Zazeil
    Nov 12 '18 at 10:30
  • 1
    $\begingroup$ @SerejaBogolubov I’ve edited my answer. I used cone the same way as you did in your question. The point is that cones (according to that definition) form a category which is isomorphic to the appropriate comma category. $\endgroup$ Nov 12 '18 at 10:50

The simplest way to think of a cone is to take any diagram, for instance: $$ \bullet \to \bullet \gets \bullet $$ where the bullets are objects in some category and the arrows are morphisms. Then you add a point "above" the diagram (the apex), with arrows pointing to each of the bullets. $$ \begin{align*} & & \color{red}\bullet & & \\ & ~~\swarrow & \downarrow & ~~~~~~\searrow & \\ \bullet & ~~~\to & \bullet & ~~~\gets & \bullet \end{align*} $$ This is a cone when every triangle coming from this construction commutes.

(You get a co-cone by adding instead a point "below", with arrows pointing from the bullets to this new point)

Thus a cone is some diagram + an apex + additional morphisms from the apex to the diagram, with some commutation relations.

Taking cones as objects and transformations of cones (which amount to transformation of one apex into another), you have a category (of cones).


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