Prove this is not a preadditive category?

[honestrosewater]

Homework Statement

Produce an example in the category in which objects are open sets in R² and morphisms are continuous maps, to illustrate that Hom sets need not be abelian groups in this category.

The Attempt at a Solution

I'm missing something stupid here. I'd think that I was looking for two open sets A,B in R² such that, for some f,g in Hom(A,B), fg != gf. No? I have no clue what the group operation is supposed to be.

 

[micromass]

Well, the hom-sets in a pre-additive categiry must be abelian groups. So, the should have a zero element. What could that zero element be?? Can there exist one??

Note that for the zero element holds that

$$f\circ 0=0=0\circ f$$

 

[honestrosewater]

$$f\circ 0=0=0\circ f$$

How can you compose two functions from the same hom-set? Am I missing something that requires the domain and codomain to be equal?

 

[micromass]

How can you compose two functions from the same hom-set? Am I missing something that requires the domain and codomain to be equal?

Let's work in the hom-set Hom(A,A) for now.

In fact, let's take A=]0,1[ and see what that gives us.

 

[honestrosewater]

The zero function z sends all points to the same point, say, r? So composition is only commutative if every other function also sends r to r? I feel totally lost. But then z(a) = 1/2 and f(a) = x/2 would work as examples for your A?

 

[micromass]

The zero function does not necessarily send every function to the same point. g is a zero function if gf=g=fg.

Now, let A=]0,1[ and let $f(x)=x^2$. What possibilities do we have for g?

 

[honestrosewater]

None?