a problem on composition of functions












3












$begingroup$


Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.



I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.










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  • $begingroup$
    Welcome! What have you tried?
    $endgroup$
    – user458276
    5 hours ago










  • $begingroup$
    Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
    $endgroup$
    – frabala
    4 hours ago












  • $begingroup$
    thank you.. got it..
    $endgroup$
    – user649511
    4 hours ago










  • $begingroup$
    What are the definitions? What does it mean to be onto?
    $endgroup$
    – JavaMan
    4 hours ago










  • $begingroup$
    onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
    $endgroup$
    – user649511
    4 hours ago
















3












$begingroup$


Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.



I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.










share|cite|improve this question









New contributor




user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Welcome! What have you tried?
    $endgroup$
    – user458276
    5 hours ago










  • $begingroup$
    Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
    $endgroup$
    – frabala
    4 hours ago












  • $begingroup$
    thank you.. got it..
    $endgroup$
    – user649511
    4 hours ago










  • $begingroup$
    What are the definitions? What does it mean to be onto?
    $endgroup$
    – JavaMan
    4 hours ago










  • $begingroup$
    onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
    $endgroup$
    – user649511
    4 hours ago














3












3








3


1



$begingroup$


Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.



I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.










share|cite|improve this question









New contributor




user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.



I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.







functions






share|cite|improve this question









New contributor




user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 33 mins ago







user649511













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asked 5 hours ago









user649511user649511

234




234




New contributor




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New contributor





user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Welcome! What have you tried?
    $endgroup$
    – user458276
    5 hours ago










  • $begingroup$
    Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
    $endgroup$
    – frabala
    4 hours ago












  • $begingroup$
    thank you.. got it..
    $endgroup$
    – user649511
    4 hours ago










  • $begingroup$
    What are the definitions? What does it mean to be onto?
    $endgroup$
    – JavaMan
    4 hours ago










  • $begingroup$
    onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
    $endgroup$
    – user649511
    4 hours ago


















  • $begingroup$
    Welcome! What have you tried?
    $endgroup$
    – user458276
    5 hours ago










  • $begingroup$
    Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
    $endgroup$
    – frabala
    4 hours ago












  • $begingroup$
    thank you.. got it..
    $endgroup$
    – user649511
    4 hours ago










  • $begingroup$
    What are the definitions? What does it mean to be onto?
    $endgroup$
    – JavaMan
    4 hours ago










  • $begingroup$
    onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
    $endgroup$
    – user649511
    4 hours ago
















$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
5 hours ago




$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
5 hours ago












$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago






$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago














$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago




$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago












$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago




$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago












$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago




$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.



Suppose $f$ is not surjective, and let



$B = f(A) tag 1$



be the image of $A$ under $f$; since



$f circ f = f, tag 2$



it is clear that every element of $B$ is fixed under $f$, for



$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$



furthermore, for $c in A$,



$f(c) = c Longrightarrow c in B; tag 4$



thus $B$ is precisely the set of fixed points of $f$.



We have assumed $f$ not surjective; then by the above we have



$B subsetneq A, tag 5$



which implies



$exists a in A setminus B; tag 6$



if



$b = f(a) in B, tag 7$



then



$f(b) = f(f(a)) = f(a) = b; tag 8$



we note



$B ni b ne a in A setminus B, tag 9$



which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    We know
    $fcirc f=f$ so $f([f(x)]) =f (x)$.



    Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
    so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...



    i see someone downvoted the question



    if you feel its a dumb question please know I'm still in school learning relations and functions ! :)






    share|cite|improve this answer










    New contributor




    user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
      $endgroup$
      – Graham Kemp
      4 hours ago



















    0












    $begingroup$

    Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.



    QED






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.



      Suppose $f$ is not surjective, and let



      $B = f(A) tag 1$



      be the image of $A$ under $f$; since



      $f circ f = f, tag 2$



      it is clear that every element of $B$ is fixed under $f$, for



      $b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$



      furthermore, for $c in A$,



      $f(c) = c Longrightarrow c in B; tag 4$



      thus $B$ is precisely the set of fixed points of $f$.



      We have assumed $f$ not surjective; then by the above we have



      $B subsetneq A, tag 5$



      which implies



      $exists a in A setminus B; tag 6$



      if



      $b = f(a) in B, tag 7$



      then



      $f(b) = f(f(a)) = f(a) = b; tag 8$



      we note



      $B ni b ne a in A setminus B, tag 9$



      which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.



        Suppose $f$ is not surjective, and let



        $B = f(A) tag 1$



        be the image of $A$ under $f$; since



        $f circ f = f, tag 2$



        it is clear that every element of $B$ is fixed under $f$, for



        $b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$



        furthermore, for $c in A$,



        $f(c) = c Longrightarrow c in B; tag 4$



        thus $B$ is precisely the set of fixed points of $f$.



        We have assumed $f$ not surjective; then by the above we have



        $B subsetneq A, tag 5$



        which implies



        $exists a in A setminus B; tag 6$



        if



        $b = f(a) in B, tag 7$



        then



        $f(b) = f(f(a)) = f(a) = b; tag 8$



        we note



        $B ni b ne a in A setminus B, tag 9$



        which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.



          Suppose $f$ is not surjective, and let



          $B = f(A) tag 1$



          be the image of $A$ under $f$; since



          $f circ f = f, tag 2$



          it is clear that every element of $B$ is fixed under $f$, for



          $b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$



          furthermore, for $c in A$,



          $f(c) = c Longrightarrow c in B; tag 4$



          thus $B$ is precisely the set of fixed points of $f$.



          We have assumed $f$ not surjective; then by the above we have



          $B subsetneq A, tag 5$



          which implies



          $exists a in A setminus B; tag 6$



          if



          $b = f(a) in B, tag 7$



          then



          $f(b) = f(f(a)) = f(a) = b; tag 8$



          we note



          $B ni b ne a in A setminus B, tag 9$



          which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.






          share|cite|improve this answer











          $endgroup$



          What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.



          Suppose $f$ is not surjective, and let



          $B = f(A) tag 1$



          be the image of $A$ under $f$; since



          $f circ f = f, tag 2$



          it is clear that every element of $B$ is fixed under $f$, for



          $b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$



          furthermore, for $c in A$,



          $f(c) = c Longrightarrow c in B; tag 4$



          thus $B$ is precisely the set of fixed points of $f$.



          We have assumed $f$ not surjective; then by the above we have



          $B subsetneq A, tag 5$



          which implies



          $exists a in A setminus B; tag 6$



          if



          $b = f(a) in B, tag 7$



          then



          $f(b) = f(f(a)) = f(a) = b; tag 8$



          we note



          $B ni b ne a in A setminus B, tag 9$



          which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          Robert LewisRobert Lewis

          47.4k23067




          47.4k23067























              0












              $begingroup$

              We know
              $fcirc f=f$ so $f([f(x)]) =f (x)$.



              Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
              so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...



              i see someone downvoted the question



              if you feel its a dumb question please know I'm still in school learning relations and functions ! :)






              share|cite|improve this answer










              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$













              • $begingroup$
                Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
                $endgroup$
                – Graham Kemp
                4 hours ago
















              0












              $begingroup$

              We know
              $fcirc f=f$ so $f([f(x)]) =f (x)$.



              Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
              so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...



              i see someone downvoted the question



              if you feel its a dumb question please know I'm still in school learning relations and functions ! :)






              share|cite|improve this answer










              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$













              • $begingroup$
                Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
                $endgroup$
                – Graham Kemp
                4 hours ago














              0












              0








              0





              $begingroup$

              We know
              $fcirc f=f$ so $f([f(x)]) =f (x)$.



              Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
              so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...



              i see someone downvoted the question



              if you feel its a dumb question please know I'm still in school learning relations and functions ! :)






              share|cite|improve this answer










              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$



              We know
              $fcirc f=f$ so $f([f(x)]) =f (x)$.



              Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
              so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...



              i see someone downvoted the question



              if you feel its a dumb question please know I'm still in school learning relations and functions ! :)







              share|cite|improve this answer










              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              share|cite|improve this answer



              share|cite|improve this answer








              edited 4 hours ago









              Graham Kemp

              86.2k43478




              86.2k43478






              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.









              answered 4 hours ago









              user649511user649511

              234




              234




              New contributor




              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              New contributor





              user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              • $begingroup$
                Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
                $endgroup$
                – Graham Kemp
                4 hours ago


















              • $begingroup$
                Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
                $endgroup$
                – Graham Kemp
                4 hours ago
















              $begingroup$
              Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
              $endgroup$
              – Graham Kemp
              4 hours ago




              $begingroup$
              Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
              $endgroup$
              – Graham Kemp
              4 hours ago











              0












              $begingroup$

              Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.



              QED






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.



                QED






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.



                  QED






                  share|cite|improve this answer









                  $endgroup$



                  Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.



                  QED







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Dbchatto67Dbchatto67

                  1,301219




                  1,301219






















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