Conclusion from the factorization theorem of functions.

[estra]

Homework Statement

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.

Homework 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..

 

[Mark44]

Homework Statement

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.

Homework 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..

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:

$$\forall$$ 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.

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.

 

[estra]

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

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.

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.

 

[Mark44]

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

 

[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..