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Math Help - Composite function

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    Exclamation Composite function

    Let A,B and C be sets and let f: A->B and g: B->c be functions. the composite function denoted by g "o" f is a function from A to C defined by
    g "o" f(x)= g(f(x)) for every x in A. Prove that if g "o" f is one to one, the f is one to one. Then prove that if g "o" f is onto, then g onto...
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nikie1o2 View Post
    Let A,B and C be sets and let f: A->B and g: B->c be functions. the composite function denoted by g "o" f is a function from A to C defined by
    g "o" f(x)= g(f(x)) for every x in A. Prove that if g "o" f is one to one, the f is one to one. Then prove that if g "o" f is onto, then g onto...
    Suppose gf:A\mapsto C is one to one but f is not. Then there exists some x,y\in A such that f(x)=f(y) but x\ne y and so g(f(x))=g(f(y)) and x\ne y. Contradiction.

    Try the other one yourself.
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    Quote Originally Posted by Drexel28 View Post
    Suppose gf:A\mapsto C is one to one but f is not. Then there exists some x,y\in A such that f(x)=f(y) but x\ne y and so g(f(x))=g(f(y)) and x\ne y. Contradiction.

    Try the other one yourself.
    In your hypothesis, what does "gf" mean, the composite? And it's a contradiction because f is one to one ??

    Do i approach the onto problem using a proof by contradiction also?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Suppose gf:A\mapsto C is one to one but f is not. Then there exists some x,y\in A such that f(x)=f(y) but x\ne y and so g(f(x))=g(f(y)) and x\ne y. Contradiction.

    Try the other one yourself.
    Quote Originally Posted by nikie1o2 View Post
    In your hypothesis, what does "gf" mean, the composite? And it's a contradiction because f is one to one ??

    Do i approach the onto problem using a proof by contradiction also?
    Yes, it's the composite function. And no, it's a contradiction because assuming that f is non-injective implies that gf is non-injective.

    For the other case it may be easier to approach it this way.

    Let c\in C then there exists some x\in A such that g(f(x))=c (since gf is surjective)...but f(x)\in B...so
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