Ulf Norell skrev:
On Jan 16, 2007, at 7:22 PM, David House wrote:
In the section on the category laws you say that the identity morphism
should satisfy
f . idA = idB . f
This is not strong enough. You need
f . idA = f = idB . f
(I do not know category theory, but try to learn from the
tutorial/article/introduction.)
Given this, and looking at the figure accompanying exercise 2. Can I not
then show that f=h from:
f.g = idA (f.g is A -> A and idA is the only morphism A -> A, closedness
gives the equality)
h.g = idA (same argument)
g.f = g.h = idB (same argument)
thus (using the laws for id and associativity)
f = idA . f = (h . g) . f = h . (g . f) = h . idB = h
Thus in the figure f=h must hold, nad one arrow can be removed from the
graph.
Regards
johan
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