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