Prove the inverse relation $f^{-1}$ of a function $f : Arightarrow B$ is a function from B to A if and only...












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This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?



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proof-explanation relations inverse-function







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  • "how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
    – fleablood
    Nov 26 at 1:50
















0














This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?



enter image description here






proof-explanation relations inverse-function







share|cite|improve this question






















  • "how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
    – fleablood
    Nov 26 at 1:50














0












0








0







This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?



enter image description here






proof-explanation relations inverse-function







share|cite|improve this question













This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?



enter image description here






proof-explanation relations inverse-function




proof-explanation relations inverse-function






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asked Nov 26 at 1:45









Nicholas Cousar

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  • "how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
    – fleablood
    Nov 26 at 1:50


















  • "how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
    – fleablood
    Nov 26 at 1:50
















"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50




"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50










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It's actually trivial.



The definition of $f^{-1}:B to A$ being surjective is:




$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.




And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.



So $f^{-1}(f(a)) = a$.



Hence $f^{-1}$ is surjective.



====




how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?




$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.



So by fact 1) $f(a) = b$.



And by fact 2) we have $(b,a) in f^{-1}$.



ANd by fact 1) again, that means $f^{-1}(b) =a$.



So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.



And that is exactly the definition of surjective.






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    1 Answer
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    It's actually trivial.



    The definition of $f^{-1}:B to A$ being surjective is:




    $f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.




    And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.



    So $f^{-1}(f(a)) = a$.



    Hence $f^{-1}$ is surjective.



    ====




    how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?




    $f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.



    So by fact 1) $f(a) = b$.



    And by fact 2) we have $(b,a) in f^{-1}$.



    ANd by fact 1) again, that means $f^{-1}(b) =a$.



    So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.



    And that is exactly the definition of surjective.






    share|cite|improve this answer




























      1














      It's actually trivial.



      The definition of $f^{-1}:B to A$ being surjective is:




      $f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.




      And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.



      So $f^{-1}(f(a)) = a$.



      Hence $f^{-1}$ is surjective.



      ====




      how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?




      $f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.



      So by fact 1) $f(a) = b$.



      And by fact 2) we have $(b,a) in f^{-1}$.



      ANd by fact 1) again, that means $f^{-1}(b) =a$.



      So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.



      And that is exactly the definition of surjective.






      share|cite|improve this answer


























        1












        1








        1






        It's actually trivial.



        The definition of $f^{-1}:B to A$ being surjective is:




        $f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.




        And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.



        So $f^{-1}(f(a)) = a$.



        Hence $f^{-1}$ is surjective.



        ====




        how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?




        $f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.



        So by fact 1) $f(a) = b$.



        And by fact 2) we have $(b,a) in f^{-1}$.



        ANd by fact 1) again, that means $f^{-1}(b) =a$.



        So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.



        And that is exactly the definition of surjective.






        share|cite|improve this answer














        It's actually trivial.



        The definition of $f^{-1}:B to A$ being surjective is:




        $f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.




        And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.



        So $f^{-1}(f(a)) = a$.



        Hence $f^{-1}$ is surjective.



        ====




        how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?




        $f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.



        So by fact 1) $f(a) = b$.



        And by fact 2) we have $(b,a) in f^{-1}$.



        ANd by fact 1) again, that means $f^{-1}(b) =a$.



        So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.



        And that is exactly the definition of surjective.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 at 2:06

























        answered Nov 26 at 1:53









        fleablood

        68k22684




        68k22684






























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