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Cancellation law with surjective functions

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data

    suppose function [tex]f : A \to B[/tex], [tex]g: A \to B[/tex], [tex]h : B \to C[/tex] satisfy [tex]g \circ f=h \circ f[/tex]. If is surjective then prove that [tex]g=h[/tex]

    2. Relevant equations


    3. The attempt at a solution

    so for any [tex]x \in A[/tex], [tex]gf(x)=hf(x)[/tex], and for any [tex]b \in B[/tex] there exist [tex]a \in A[/tex], such that [tex]f(b)=a[/tex]

    so [tex]g(a)=h(a)[/tex] so [tex]g=h[/tex] is this correct and sufficient?

    i'm suppose to to show for any [tex]v \in B[/tex], [tex]g(v)=h(v)[/tex]. i don't know but something's missing.
    Last edited: Sep 9, 2010
  2. jcsd
  3. Sep 9, 2010 #2


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    Re: function

    You need to clean it up a bit. For one thing, you have g mapping A to B. You meant B to C, right? Also, the element a in in A, but h is defined on B, so when you say h(a), it's not clear that h(a) is defined.
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