Is it sufficient to consider control operators?

[tomjaguarpaw]

Perhaps we don't have to consider arrows as separate from control operators. Given f :: arr a b we can write e = (>>> f) :: arr e a -> arr e b. Following definition 4 (page 5) of "A new notation for arrows" we need to check that "e satisfies a ... naturality property"

e ((k ~> θ1) x1) ... ((k ~> θn) xn) = (k ~> θ) (e x1 ... xn)

For our e this amounts to showing that e ((arr k >>>) x) = (arr k >>>) (e x) which indeed holds. Thus every arrow gives rise to a control operator. On the other hand we can recover the arrow from the control operator: f = e id.

Therefore it seems that considering control operators only is sufficient.

[tomjaguarpaw]

Oh, wait. f is also a control operator via arr snd >>> f 😄