Re: [Haskell-beginners] Category question

On Mon, May 28, 2012 at 02:08:14PM +0200, Manfred Lotz wrote:

On Mon, 28 May 2012 13:34:01 +0200 Alessandro Pezzoni wrote:

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On Mon, May 28, 2012 at 11:54:11AM +0200, Manfred Lotz wrote:

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In the definition of a (mathematical) category it is said (among other things), that for any object A there exists an identity morphism:

idA: A -> A and if f: A -> B for two objects A, B then

idB . f = f and f . idA = f

must hold.

My question: Because I cannot think of any counterexample for the last statement I would like to know if I just could omit this from the definition and formulate this as a small theorem.

Or does there exist a counterexample where all conditions of a category hold but there exist two objects A, and B where we have idB . f <> f and/or f .idA <> f?

When you ask that idB . f = f and f . idA = f you are basically defining a left and a right identity, respectively.

If I get your question correctly, you are asking if you can drop the axiom (requirement) of existence of an identity morphism for every object and deduce it from the other axioms, i.e. that the composition of morphisms is always well defined and that it is associative.

No, I do not want to drop the requirement of existence of an identity morphism. I only want to drop the axion that idB .f = f and f . idA = f do hold because IMHO this follows readily from the definition of what an identity morphism is all about.

"Follows readily from the definition of what an identity morphism is all about" -- and what exactly is that defintion? In fact, the definition is precisely that idB . f = f and f . idA = f This is not an "extra" requirement on identity morphisms. It is simply the definition. -Brent