# Prove this is not a preadditive category?

1. Sep 12, 2011

### honestrosewater

1. The problem statement, all variables and given/known data

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.

3. 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.

2. Sep 12, 2011

### 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

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

3. Sep 12, 2011

### honestrosewater

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

4. Sep 12, 2011

### micromass

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.

5. Sep 12, 2011

### 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?

6. Sep 12, 2011

### 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???

7. Sep 12, 2011

None?