Re: [Haskell-cafe] adjunction of product and exponentiation

7 Aug 2008

      On Wed, 2008-08-06 at 22:50 -0700, Jason Dusek wrote:
...
Jonathan Cast  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)?
curry f x y = f (x, y)

So therefore

  curry id x y
= id (x, y)
= (x, y)

Eta-contracting, we get

curry id = \ x. \ y. (x, y)

which is your function.

jcc