# Conclusion from the factorization theorem of functions.

1. Oct 25, 2008

### estra

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

Prove that $$\forall$$ f:X$$\rightarrow$$Y there $$\exists$$ Z, h: X$$\rightarrow$$Z is injective and g: Z$$\rightarrow$$Y is surjective, so that f=g*h.

2. Relevant equations

There is already a conclusion from the factorisation theorem of functions that: $$\forall$$ f:X$$\rightarrow$$Y there $$\exists$$ Z, h: X$$\rightarrow$$Z is surjective and g: Z$$\rightarrow$$Y is injective, so that f=g*h.

But how to prove it just from applaying it from other side..

2. Oct 25, 2008

### Staff: Mentor

What do you mean "applaying it from the other side.."?

I for one need some context here. In this equation, f=g*h, '*' looks like ordinary multiplication, but with the sets that are involved, do you mean composition (i.e., g $$\circ$$ h)?

Also, it's not clear to me what you're saying here:
I don't understand the stuff above. What I think you are saying is:
For any f, where f:X$$\rightarrow$$Y, there is an injective function h, where h: X$$\rightarrow$$Z, and a surjective function g, where g: Z$$\rightarrow$$Y, so that f = g$$\circ$$h.

3. Oct 26, 2008

### estra

Ok. Don't watch the stuff above then. I might have sentenced it wrong.

You are absoultely right. This is what I try to say and proof. Yes I mean composition. Sorry about the wrong symbol. g $$\circ$$ h is what i meant.

4. Oct 26, 2008

### Staff: Mentor

So what did you mean by "applaying it from the other side.."?

5. Oct 26, 2008

### estra

I just tried to say that almost the same sentence (only when function h is surjective and function g is injective) is a conclusion from a theorem(functions factorization theorem). but i try to proof the sentence where h is injective and g is surjective.. so i thought maybe it can be proved on the same way..