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    • CommentRowNumber1.
    • CommentAuthorbblfish
    • CommentTimeAug 22nd 2014
    • (edited Aug 22nd 2014)
    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?
    • CommentRowNumber2.
    • CommentAuthorZhen Lin
    • CommentTimeAug 22nd 2014

    There is a natural bijection between functors ICI \to C and morphisms in CC, namely the map that sends a functor ICI \to C to the image in CC of the morphism 010 \to 1 in II.

    • CommentRowNumber3.
    • CommentAuthorTobyBartels
    • CommentTimeAug 22nd 2014

    Funct(I,C)Funct(I,C) is not itself a map from II to CC; it is a category (the arrow category itself) whose objects are maps (functors) from II to CC. So an object of Funct(I,C)Funct(I,C) is a functor FF from II to CC.

    And what does this functor F:ICF\colon I \to C consist of? II has two objects, 00 and 11; so we need two objects of CC, F(0)F(0) and F(1)F(1). Similarly, II has three morphisms, id 0:00id_0\colon 0 \to 0, i:01i\colon 0 \to 1, and id 1:11id_1\colon 1 \to 1; so we need three morphisms of CC, F(id 0):F(0)F(0)F(\id_0)\colon F(0) \to F(0), F(i):F(0)F(1)F(i)\colon F(0) \to F(1), and F(id 1):F(1)F(1)F(\id_1)\colon F(1) \to F(1). Since the functor FF must preserve identities, we need F(id 0)=id F(0)F(\id_0) = \id_{F(0)} and F(id 1)=id F(1)F(\id_1) = \id_{F(1)}, so the data remaining is precisely that in the morphism f(i):F(0)F(1)f(i)\colon F(0) \to F(1). It now remains to check that the functor FF preserves composition, but it does.

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

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

    • CommentRowNumber4.
    • CommentAuthorbblfish
    • CommentTimeAug 23rd 2014
    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.
    • CommentRowNumber5.
    • CommentAuthorTobyBartels
    • CommentTimeAug 23rd 2014

    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.

    • CommentRowNumber6.
    • CommentAuthorbblfish
    • CommentTimeAug 23rd 2014

    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.