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Tricky algebra proof

  1. Jan 10, 2006 #1
    tricky algebra proof....


    consider the mappings: (R: A-->B)

    Suppose T: C-->A and S: C-->A satisfy RT = RS then T = S
    prove that there exists a; b: B-->A such that bR = idA (identity of A)

    well, im not sure if I can say that b (inverse) = R
    since b maps B to A .. and R maps A to B.... and if i can say that...
    how do i approach the proof?

    I know that RT = RS then T = S.. should i work with this.? using b (inverse)...
    or should i try to see if R or b is injective?....

    ie. b(inverse)T = b(inverse)S ... ??
  2. jcsd
  3. Jan 11, 2006 #2

    matt grime

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    Science Advisor
    Homework Helper

    The idea of left cancellable (RS=RT => S=T) is exactly the same as R being injective.
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