Exercise 4.1.31

Theorem: Let be the functor sending a small category to the set of all its maps.

the category with two objects and one non-identity arrow between them. We will denote the objects by and and the map by . We claim . Indeed, we define a natural transformation . For a category , define as follows. For a map in , let be the obvious functor sending to , to , and the to . To check naturality, suppose we have a functor . We must have . Note that is essentially (the restriction of to the maps of the category), and is left-composition by . Start with a map in . It's straightforward to see that applying the left and right side result in the same functor . Applying the right side to yields:</script>

f \enspace \mapsto \enspace \left{ \begin{array}{l} \circ \mapsto A \

  • \mapsto A' \ m \mapsto f \end{array} \right. \enspace \mapsto \enspace \left{ \begin{array}{l} \circ \mapsto F(A) \
  • \mapsto F(A') \ m \mapsto F(f) \end{array} \right.

    $$</li> </ul>

    while applying the left side to yields:

    so is natural and .