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Math Help - Functions and Inverses

  1. #1
    Senior Member tukeywilliams's Avatar
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    Functions and Inverses

    I am not exactly sure what these functions mean:

     \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y) defined by  \overrightarrow{f}(A) = \{ f(x) | x \in A \} for  A \in \mathcal{P}(X) and

     \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X) defined by  \overleftarrow{f}(B) = \{x \in X | f(x) \in B \} for  B \in \mathcal{P}(Y) .

    These two functions really are  f and  f^{-1} right?

    So if  f is a bijection with inverse  f^{-1} , then  f(x) = y_0 if and only if  x = f^{-1}(y_0) so that  \overleftarrow{f}(\{y_0 \}) = \{f^{-1}(y_0) \} . So this implies that  \overleftarrow{f}(\{y_0 \}) \neq \{f^{-1}(y_0) \} sometimes?
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    Quote Originally Posted by tukeywilliams View Post
     \overrightarrow{f}: \mathcal{P}(X) \rightarrow \mathcal{P}(Y) defined by  \overrightarrow{f}(A) = \{ f(x) | x \in A \} for  A \in \mathcal{P}(X) and

     \overleftarrow{f}: \mathcal{P}(Y) \rightarrow \mathcal{P}(X) defined by  \overleftarrow{f}(B) = \{x \in X | f(x) \in B \} .

    These two functions really are  f and  f^{-1} right?
    No, they are really two new functions related to the function  f .
    The two new functions are mapping of subsets to subsets.
    Whereas  f maps set X to set Y.
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  3. #3
    Senior Member tukeywilliams's Avatar
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    Yes, but why do we need these functions? Are they only useful for figuring out whether functions have inverses?

    In other words, if  \overleftarrow{f}(\{y_0\}) \neq \{f^{-1}(y_0)\} then  f has no inverse?
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    Quote Originally Posted by tukeywilliams View Post
    Yes, but why do we need these functions?
    This inverse image function is really useful.

    I do not think you taken group theory, given the homomorphism \phi: G \mapsto G' then \phi ^{-1} [\{e\}] where \{e'\}\subseteq G' is called the kernel it is extremely important concept.

    Have you even taken linear algebra? Given a linear transformation T: V\to V' then \ker T is defined exactly like above T^{-1} [ \bold{0}] where \bold{0} is the zero vector of V'. This is another important concept.
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    Senior Member tukeywilliams's Avatar
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    Yes I have taken linear algebra.
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    Quote Originally Posted by tukeywilliams View Post
    Yes I have taken linear algebra.
    Do you understand what I mean by the kernel of a linear transformation. Carefully try to understand that the kernel is defined as an inverse image.
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  7. #7
    Senior Member tukeywilliams's Avatar
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    Yeah the kernel of a linear transformation is the set of vectors v such that  T(v) =0 , i.e. the set of vectors that maps to the 0 vector.

    So the inverse image function takes the kernel and maps it to the singleton set  \{0 \} and vice versa.
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    Quote Originally Posted by tukeywilliams View Post
    Yes, but why do we need these functions?

    In other words, if  \overleftarrow{f}(\{y_0\}) \neq \{f^{-1}(y_0)\} then  f has no inverse?
    The answer to the first question is: it is a foundational issue. That is, in a course on the foundation of mathematics, logic & sets, a function is defined as a relation with additional properties. If f is a function from A to B then f is also a relation from A to B. Now it is true then that f^{-1} is also a relation from A to B BUT it may not be a function.

    However \overleftarrow f is always a function from P(B) to P(A).
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