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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  • "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















up vote
0
down vote

favorite












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










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













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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










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?



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proof-explanation relations inverse-function






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asked Nov 26 at 1:45









Nicholas Cousar

302211




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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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    1 Answer
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    up vote
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    down vote



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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



























      up vote
      1
      down vote



      accepted










      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

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        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

        67k22684




        67k22684






























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