If we are talking about comment #3 then you are right Simon.

But if we ignore the fact that Dan didn't choose one over the other, then presumably John preferred the version with \\(f \circ f = 1\_\star\\) because then f has an inverse (and hence is then a group).

But if we ignore the fact that Dan didn't choose one over the other, then presumably John preferred the version with \\(f \circ f = 1\_\star\\) because then f has an inverse (and hence is then a group).