Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

1. • CommentRowNumber1.
   • CommentAuthorbblfish
   • CommentTimeAug 22nd 2014
   • (edited Aug 22nd 2014)
   • PermaLink
   Author: bblfish
   Format: TextThe Arrow Category page says "For any category C, its arrow category is the functor category Arr(C):=Funct(I,C) for I the interval category {0→1}" ... "This means that the objects of Arr(C) are the morphisms of C" But Funct(I,C) should be a map between I towards C, which cant be an object whose morphisms are C. I am new to CT, and am very likely reading this wrong. But how should I read it?

   The Arrow Category page says
   "For any category C, its arrow category is the functor category Arr(C):=Funct(I,C) for I the interval category {0→1}"
   ...
   "This means that the objects of Arr(C) are the morphisms of C"
   
   But Funct(I,C) should be a map between I towards C, which cant be an object whose morphisms are C.
   
   I am new to CT, and am very likely reading this wrong. But how should I read it?
2. • CommentRowNumber2.
   • CommentAuthorZhen Lin
   • CommentTimeAug 22nd 2014
   • PermaLink
   Author: Zhen Lin
   Format: MarkdownItexThere is a natural bijection between functors $I \to C$ and morphisms in $C$, namely the map that sends a functor $I \to C$ to the image in $C$ of the morphism $0 \to 1$ in $I$.

   There is a natural bijection between functors and morphisms in , namely the map that sends a functor to the image in of the morphism in .

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeAug 22nd 2014
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex$Funct(I,C)$ is not itself a map from $I$ to $C$; it is a category (the [[arrow category]] itself) whose objects are maps (functors) from $I$ to $C$. So an object of $Funct(I,C)$ is a functor $F$ from $I$ to $C$. And what does this functor $F\colon I \to C$ consist of? $I$ has two objects, $0$ and $1$; so we need two objects of $C$, $F(0)$ and $F(1)$. Similarly, $I$ has three morphisms, $id_0\colon 0 \to 0$, $i\colon 0 \to 1$, and $id_1\colon 1 \to 1$; so we need three morphisms of $C$, $F(\id_0)\colon F(0) \to F(0)$, $F(i)\colon F(0) \to F(1)$, and $F(\id_1)\colon F(1) \to F(1)$. Since the functor $F$ must preserve identities, we need $F(\id_0) = \id_{F(0)}$ and $F(\id_1) = \id_{F(1)}$, so the data remaining is precisely that in the morphism $f(i)\colon F(0) \to F(1)$. It now remains to check that the functor $F$ preserves composition, but it does. So a functor from $I$ to $C$ consists of precisely the same information as a morphism in $C$. We still have to check that the morphisms in $Funct(I,C)$ (which are natural transformations) correspond in this way to commutative squares in $C$, as claimed; I leave that to you.

   is not itself a map from to ; it is a category (the arrow category itself) whose objects are maps (functors) from to . So an object of is a functor from to .

   And what does this functor consist of? has two objects, and ; so we need two objects of , and . Similarly, has three morphisms, , , and ; so we need three morphisms of , , , and . Since the functor must preserve identities, we need and , so the data remaining is precisely that in the morphism . It now remains to check that the functor preserves composition, but it does.

   So a functor from to consists of precisely the same information as a morphism in .

   We still have to check that the morphisms in (which are natural transformations) correspond in this way to commutative squares in , as claimed; I leave that to you.

4. • CommentRowNumber4.
   • CommentAuthorbblfish
   • CommentTimeAug 23rd 2014
   • PermaLink
   Author: bblfish
   Format: TextThanks @TobyBartels, that helped a lot. There should be links from the wiki page to relevant answers like that, as it is very useful. I drew a few diagrams out and convinced myself it works.

   Thanks @TobyBartels, that helped a lot. There should be links from the wiki page to relevant answers like that, as it is very useful. I drew a few diagrams out and convinced myself it works.
5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeAug 23rd 2014
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexOne problem with this article is that it still shows our early tendency to give everything the slickest definition possible. I\'ve rewritten it in a way that I think is more comprehensible.

   One problem with this article is that it still shows our early tendency to give everything the slickest definition possible. I've rewritten it in a way that I think is more comprehensible.

6. • CommentRowNumber6.
   • CommentAuthorbblfish
   • CommentTimeAug 23rd 2014
   • PermaLink
   Author: bblfish
   Format: MarkdownItexyes, I see. While the previous definition is good for the brain if one has faith enough to pursue it, it does leave one with the feeling that in CT the easy is explained by the more complex. With your new definition the intuitive is explained first, then one gets an identity statement with a purer, though more complex way of stating things.

   yes, I see. While the previous definition is good for the brain if one has faith enough to pursue it, it does leave one with the feeling that in CT the easy is explained by the more complex. With your new definition the intuitive is explained first, then one gets an identity statement with a purer, though more complex way of stating things.