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Thread: Sets and functions

  1. #1
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    Sets and functions

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

    $\displaystyle 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} $

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

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

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

    (2) Suppose that $\displaystyle f: X \to Y $ and $\displaystyle g: Y \to Z $ are surjections. Prove that the composite $\displaystyle g \circ f: X \to Z $ is also a surjection. So I started out by saying that $\displaystyle \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 $\displaystyle \forall z \in Z, \ gf(x) = z $?

    (3) Let $\displaystyle f: X \to Y $ be a function. Prove that there exists a function $\displaystyle g: Y \to X $ such that $\displaystyle f \circ g = I_{y} $ if and only if $\displaystyle f $ is a surjection. So we want to prove the following: $\displaystyle f \circ g = I_{y} \Leftrightarrow $ $\displaystyle 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 $\displaystyle f(g(x)) = y $, $\displaystyle g(x) $ points back to the domain of $\displaystyle f $. Then $\displaystyle \forall x \in X, \ y = f(x) $. Is this correct?

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

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


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

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