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

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

On Mon, 28 May 2012 10:04:33 -0400 Brent Yorgey wrote:

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:

...

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.

"Follows readily from the definition of what an identity morphism is all about" -- and what exactly is that defintion?

For me id: A -> A could be defined by: A morphism id: A -> A is called identity morphism iff for all x of A we have id(x) = x. IMHO, from this the following follows readily.

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.

I agree that I could define id by these two statements. My point is that in the books about category theory those two statements are stated as axioms, and id is (in many books) just self understood or defined as I have defined it above. If in a book about category the author would say that for each object A there must exist a morphism id: A -> A (called identity morphism) which is defined by idB . f = f and f . idA = f then this would be clearer (and better, IMHO). -- Manfred -- Manfred

On Mon, May 28, 2012 at 04:14:40PM +0200, Manfred Lotz wrote:

For me id: A -> A could be defined by: A morphism id: A -> A is called identity morphism iff for all x of A we have id(x) = x.

This is not actually a valid definition; the notation id(x) = x does not make sense. It seems you are assuming that morphisms represent some sort of function, but that is only true in certain special categories.

My point is that in the books about category theory those two statements are stated as axioms, and id is (in many books) just self understood or defined as I have defined it above.

If in a book about category the author would say that for each object A there must exist a morphism id: A -> A (called identity morphism) which is defined by idB . f = f and f . idA = f then this would be clearer (and better, IMHO).

This is exactly what category theory books do (or should) say. Do you have a particular example of a book which does not state things in this way? Note that there is no particular difference between calling these equations "axioms" or a "definition". That is, "there is an 'identity morphism' satisfying the following axioms..." and "there is an 'identity morphism' defined by..." are just two different ways of saying the exact same thing. -Brent