@Jonathan

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

\$$\mathrm{FriendOf}(\alpha(\textrm{Alice})) = \alpha(\mathrm{FriendOf}(\textrm{Alice})) = \textrm{Bob}\$$

There is also the transformation where you collapse everything down to one dot : Alice to Bob and Tyler to Bob which is trivial.