On Mon, 28 May 2012 13:34:01 +0200
Alessandro Pezzoni
On Mon, May 28, 2012 at 11:54:11AM +0200, Manfred Lotz wrote:
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. Or in other words: I do not understand why all definitions of a category say that for all objects an identity morphism has to exist and then they say that those id morphism have to hold above conditions instead of saying that those conditions follows directly from the definition of id. -- Manfred
This is not possible, though. As a simple example, consider a category with only one object, let's call it X, and a non identical morphism f: X -> X which is such that f^n = f . ... . f is not identical for every non-zero natural n. For example you can consider X as a totally ordered infinite set, e.g. the set of integers Z, and f the map that shifts every element by one to the right, e.g. f(z) = z + 1 for every z in Z, so that f^n(z) = z + n != z for every non-zero natural n. If now you consider the "category" with only X as object and as set of morphisms {f^n | n is a non-zero natural}, then the composition is always defined as f^n . f^m = f^(n+m) and it is obviously associative. But in this example there is no identical morphism, because f^n(z) != z for every non-zero natural n.
By the way, this example models a semigroup, like the set of natural numbers N\{0} with the canonical sum.
Alessandro
-- Manfred