Michael Hong wrote:

> **Puzzle 128** What about projecting Alice onto Bob? Wouldn't this commute as well?

I don't believe so. With this transformation, \\(\alpha(\textrm{FriendOf}(\textrm{Bob})) = \alpha(\textrm{Alice}) = \textrm{Bob}\\), but \\(\textrm{FriendOf}(\alpha(\textrm{Bob})) = \textrm{FriendOf}(\textrm{Bob}) = \textrm{Alice}\\).

The same reasoning denies the collapse-everything transformation. The key here is that we're using the same \\(\textrm{FriendOf}\\) function either before or after applying the natural transformation, meaning we can easily escape the "collapse" after following one of the natural transformation arrows.

Ryan Wisnesky wrote:

> Hi Jonathan (#3), it is not the case that the image of a functor F : C -> D is a subcategory of D. See https://math.stackexchange.com/questions/413138/can-it-happen-that-the-image-of-a-functor-is-not-a-category

Wow! I did not expect this twist. I'll have to reflect on that...

> **Puzzle 128** What about projecting Alice onto Bob? Wouldn't this commute as well?

I don't believe so. With this transformation, \\(\alpha(\textrm{FriendOf}(\textrm{Bob})) = \alpha(\textrm{Alice}) = \textrm{Bob}\\), but \\(\textrm{FriendOf}(\alpha(\textrm{Bob})) = \textrm{FriendOf}(\textrm{Bob}) = \textrm{Alice}\\).

The same reasoning denies the collapse-everything transformation. The key here is that we're using the same \\(\textrm{FriendOf}\\) function either before or after applying the natural transformation, meaning we can easily escape the "collapse" after following one of the natural transformation arrows.

Ryan Wisnesky wrote:

> Hi Jonathan (#3), it is not the case that the image of a functor F : C -> D is a subcategory of D. See https://math.stackexchange.com/questions/413138/can-it-happen-that-the-image-of-a-functor-is-not-a-category

Wow! I did not expect this twist. I'll have to reflect on that...