[Haskell-cafe] adjunction of product and exponentiation

[Jason Dusek]

Jonathan Cast <jonathanccast at fastmail.fm> wrote:
> Jason Dusek wrote:
> > It is an arrow that takes a C to an arrow that takes an A
> > and makes the product C x A. I want to write curry(C x A)
> > but that is ridiculous looking. What's the right notation
> > for this thing?
>
> It's a curried pairing operator. Haskell calls it (,); it
> might also be called pair. It is also, of course, equal to
> curry(id), so if you write identity arrows as the
> corresponding objects then curry(C x A) is perfectly
> reasonable.

  Why is it equal to curry(id)?

-- 
_jsn