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Composition of functions

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Let f: A→B and g: B→C both be bijections. Prove that the composed function fog is also a bijection. Write (fog)-1 in terms of f-1 and g-1.


    2. Relevant equations

    fog = f(g(x))

    3. The attempt at a solution

    I know what the composed function means and what bijective means (one-to-one). I can do these problems when I have values for f(x) and g(x) but I'm having trouble with this type of problem. I nod in the right direction would be much appreciated.
     
    Last edited: May 8, 2014
  2. jcsd
  3. May 8, 2014 #2

    pasmith

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    Homework Helper

    If [itex]f: A \to B[/itex] and [itex]g: B \to C[/itex] then the composition of the two which makes sense is [itex]g \circ f : A \to C : x \mapsto g(f(x))[/itex].

    A bijection is both an injection and a surjection. Have you shown that a composition of injections is an injection, and a composition of surjections is a surjection?

    Alternatively, you could exhibit an inverse of [itex]g \circ f[/itex] in terms of the inverses of [itex]g[/itex] and [itex]f[/itex].
     
  4. May 8, 2014 #3
    First of all thanks, and I made a mistake in the question which I changed ie. the composed function was a square but now an inverse.

    Secondly, the question asked for fog not gof?

    Injectivity: If x,y [itex]\in[/itex]A and g(f(x))=g(f(y)) then f(x)=f(y) and x=y

    Surjectivity: If c [itex]\in[/itex]C then g(b)=c and f(a)=b

    Hence g(f(a))=g(b)=c

    Is that sufficient to prove its a bijection?

    And how do I write Write (fog)-1 in terms of f-1 and g-1?
     
  5. May 8, 2014 #4
    Yes, the logic and solution is fine. Just justify the parts where you assume a certain property of a function, so that the reader(grader) knows that you understand what your doing at each step. So instead of saying f(x)=f(y),x=y.
    Instead say f(x)=f(y) implies x=y because the function f is bijective.
     
  6. May 11, 2014 #5
    Ok thanks for the help :)
     
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