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Math Help - Sets and functions

  1. #1
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    Sets and functions

    (1) Functions  f: \mathbb{R} \to \mathbb{R} and  g: \mathbb{R} \to \mathbb{R} are defined as follows:

     f(x) = \begin{cases} x+2\ \ \text{if}\ x<-1 \\ -x\ \ \ \text{if} -1 \leq x \leq 1 \\ x-2\ \text{if} \ \ x >1 \end{cases}

     g(x) = \begin{cases} x-2\ \ \text{if}\ x<-1 \\ -x\ \ \ \text{if} -1 \leq x \leq 1 \\ x+2\ \text{if} \ \ x >1 \end{cases}

    Find functions  f \circ g and  g \circ f . Is  g the inverse of  f ? Is  f injective or surjective? How about  g ?

    So  f \circ g = x = g \circ f (i.e. the identity element  x ).

    I said  g is the inverse function of  f . I said  f is surjective because  f(-2) = f(0) (hence not injective). I am not sure whether  g is injective or surjective.

    (2) Suppose that  f: X \to Y and  g: Y \to Z are surjections. Prove that the composite  g \circ f: X \to Z is also a surjection. So I started out by saying that  \forall y \in Y, \ f(x) = y \ \text{and} \ \forall z \in Z, \ g(y) = z . Then how would I go from this to showing that  \forall z \in Z, \ gf(x) = z ?

    (3) Let  f: X \to Y be a function. Prove that there exists a function  g: Y \to X such that  f \circ g = I_{y} if and only if  f is a surjection. So we want to prove the following:  f \circ g = I_{y} \Leftrightarrow  f is a surjection. This is equivalent to saying, from my earlier post, that a function is surjective if and only if it has a right inverse. So given that  f(g(x)) = y ,  g(x) points back to the domain of  f . Then  \forall x \in X, \ y = f(x) . Is this correct?

    Thanks
    Last edited by shilz222; August 21st 2007 at 03:04 PM.
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  2. #2
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    \begin{array}{lcr}<br />
 g(3) = 5,\;f(5) = 3\quad  &\Rightarrow & \quad f \circ g(3) = 3 \\ <br />
 f(3) = 1,\;g(1) =  - 1\quad  &\Rightarrow & \quad g \circ f(3) =  - 1 \\ <br />
 \end{array}
    I think that you need to redo the complete problem.
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  3. #3
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    Ok so  f \circ g \neq g \circ f . Since  f is not bijective its inverse is not  g .  f is surjective, and I think  g is bijective?
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  4. #4
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    Yeah I drew that graph. It looks like  g(x) is bijective.
    Last edited by shilz222; August 21st 2007 at 01:11 PM.
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  5. #5
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    Quote Originally Posted by shilz222 View Post
    Yeah I drew that graph. It looks like  g(x) is bijective.
    BIJECTIVE?
    g( - 2) = g(0) = g(2) = 0
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  6. #6
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     g(-2) = -4 . Your talking about  f(x) which is surjective.
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  7. #7
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    Quote Originally Posted by shilz222 View Post
    Your talking about  f(x) which is surjective.
    O.K.

    But is  g onto?
     g(?) = 2
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     g(x) is not surjective. Its injective.
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  9. #9
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    Quote Originally Posted by shilz222 View Post
     g(x) is not surjective. Its injective.
    Correct.
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  10. #10
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    For (2)  g(f(x)) = g(y) = g(z) \implies g \circ f is surjective. Is that correct?
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  11. #11
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    (2) Suppose that  f: X \to Y and  g: Y \to Z are surjections.
    Prove that the composite  g \circ f: X \to Z is also a surjection.


    Suppose that q \in Z then \left( {\exists p \in Y} \right)\left[ {g(p) = q} \right]. But then \left( {\exists m \in X} \right)\left[ {f(m) = p} \right]; therefore g \circ f(m) = g\left( {f\left( m \right)} \right) = g\left( p \right) = q, surjective.
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  12. #12
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    My proof was more like this: Consider  f: X \to Y and  g: Y \to Z both surjective. Given any  z \in Z we must find an  x such that  f(x) = z . Since  g is surjective there is a  y \in Y such that  g(y) = z . Since  f is surjective there is an  x \in X such that  f(x) = y . Therefore  g \circ f(x) = g(f(x)) = g(y) = g(z) .
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  13. #13
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    (3) Let  f: X \to Y be a function. Prove that there exists a function  g: Y \to X such that  f \circ g = I_{y} if and only if  f is a surjection. So we want to prove the following:  f \circ g = I_{y} \Leftrightarrow  f is a surjection. This is equivalent to saying, from my earlier post, that a function is surjective if and only if it has a right inverse. So given that  f(g(x)) = y ,  g(x) points back to the domain of  f . Then  \forall x \in X, \ y = f(x) . Is this correct?
    So given  f \circ g = I_y then  f(g(y) = f(x) = y,  \ \forall x \in X and so  f is a surjection. Given that  f is a surjection, then  \forall x \in X, \ y = f(x) . This implies that  f \circ g = I_y .

    Is this correct?
    Last edited by shilz222; August 21st 2007 at 10:12 PM.
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