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Conclusion from the factorization theorem of functions.

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    But how to prove it just from applaying it from other side..
  2. jcsd
  3. Oct 25, 2008 #2


    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 [tex] \circ [/tex] 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[tex]\rightarrow[/tex]Y, there is an injective function h, where h: X[tex]\rightarrow[/tex]Z, and a surjective function g, where g: Z[tex]\rightarrow[/tex]Y, so that f = g[tex]\circ[/tex]h.
  4. Oct 26, 2008 #3
    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 [tex] \circ [/tex] h is what i meant.
  5. Oct 26, 2008 #4


    Staff: Mentor

    So what did you mean by "applaying it from the other side.."?
  6. Oct 26, 2008 #5
    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..
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