Composites of injections/surjections/bijections

  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. StatusX

    StatusX 2,567
    Homework Helper

    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. No there aren't any other conditions.
     
  5. StatusX

    StatusX 2,567
    Homework Helper

    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. O I made a mistake in the question, we are given that h o g o f is bijective.
     
  7. StatusX

    StatusX 2,567
    Homework Helper

    Ok, then use the fact that bijections have inverses, which are also bijections, and that the composition of bijections is a bijection.
     
  8. got it, thanks!
     
  9. 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. StatusX

    StatusX 2,567
    Homework Helper

    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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