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Composites of injections/surjections/bijections

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider arbitrary sets A, B, C and D with arbitrary functions:
    f:A-->B, g:B-->C, h:C-->D. We define a composite function
    h o g o f:A-->D.
    Given that h, f, and h o g o f are bijective, and g is injective, show that g is also surjective (i.e. g is bijective).
    This seems almost trivial to me, but my TA says that it requires proving.
     
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2

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    I don't understand the question. Are there any conditions on g besides it being injective? You can always compose functions between sets like that, so that doesn't tell you anything.
     
  4. May 8, 2007 #3
    No there aren't any other conditions.
     
  5. May 8, 2007 #4

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    Here's a similar question: Let x,y,z be integers, and define x+y+z. Show that if x and z are even, so is y. Do you see why you need more information? And what made you think it was trivial?
     
  6. May 8, 2007 #5
    O I made a mistake in the question, we are given that h o g o f is bijective.
     
  7. May 8, 2007 #6

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    Ok, then use the fact that bijections have inverses, which are also bijections, and that the composition of bijections is a bijection.
     
  8. May 8, 2007 #7
    got it, thanks!
     
  9. May 8, 2007 #8
    One more question:
    In our textbook we are given a theorem that:
    If f o g is bijective then g is injective and f is surjective. I can informally see this by drawing Venn Diagrams, but how would one go about doing a formal proof.
     
  10. May 8, 2007 #9

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    Just go back to the definitions. For example, assume g wasn't injective. Then for some x,y, we'd have g(x)=g(y), and so f(g(x))=f(g(y)), and f o g isn't injective.
     
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