Recent content by 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...

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.

Homework Statement Prove that \forall f:X\rightarrowY there \exists Z, h: X\rightarrowZ is injective and g: Z\rightarrowY is surjective, so that f=g*h. Homework Equations There is already a conclusion from the factorisation theorem of functions that: \forall f:X\rightarrowY there...

well that is how it is asked in my textbook. and i was told to prove that. there is no more information about that there...

ok sorry about that. Lets say it again. Prove that, when S and R are equivalence relations in the set X, then RS is an equivalence relation in the set X when SR=RS. I think that on the second part we have to show that the composition of equivalence relations is commutative.. (when for...

But who do you what "RS" and "SR" mean when S and R are subsets of X x X I can only think that way S ∘ R = { (x, z) | there exists y ∈ Y, such that (x, y) ∈ R and (y, z) ∈ S }.

Homework Statement Prove that when R and S are equivalence relations, then SR is equivalance relation when SR=RS. Homework Equations The Attempt at a Solution I want to know what sets should i take for the relations ? is it ok when i take R a a subset of X x Y and S as a...

OK. Let's assume that R \subset X x Y is a bijection between X and Y. Then we also have to have R-1. What is R-1 ⊂ Y x X when i: X → X then (x,y)\inR has to be written (f(x),x) \in R and x \in X and also (f-1)y,y \in R and x \in X. but how to show that R-1R=I abd RR-1=I ? or do we have...

OK. Let's assume that R \subset X x Y is a bijection between X and Y. Then we also have to have R-1. What is R-1 ⊂ Y x X when i: X → X then (x,y)\inR has to be written (f(x),x) \in R and x \in X and also (f-1)y,y \in R and x \in X. but how to show that R-1R=I abd RR-1=I ? or do we have...

I only know that if R ⊂ X x Y then R-1 ⊂ Y x X or: (y,x)\inR-1 \Leftrightarrow (x,y) \in R

Prove that R ⊂ X x Y is a bijection between the sets X and Y, when R−1R= I: X→X and RR-1=I: Y→Y Set theory is a quite a new lesson for me. So I am not good at proving different connections, but please give me a little help with what to start and so.. I have read the book and I know what...