How do I determine whether a function/map T is one-to-one and/or onto?












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$begingroup$


Given $T(x,y,z) = (xy, yz, xz)$, determine whether $T$ is one to one and/or onto.



So far, I have come up with a contradiction that proves $T$ is not one to one:



$T(-1, 1, -1) = (-1, -1, 1)$
and, $T(1, -1, 1) = (-1, -1, 1)$. However, I can't figure out whether or not $T$ is onto... I think it is, but I don't know the reaosn why or how to prove it.



Can I say:



Taking arbitrary elements a,b,c $in mathbb R$, $T(a,b,c) = (ab, bc, ac)$, and therefore any element $x,y,z in mathbb R$ can be generated by $x=ab$, $y = bc$, and $z=ac$?






calculus functions linear-transformations







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    0












    $begingroup$


    Given $T(x,y,z) = (xy, yz, xz)$, determine whether $T$ is one to one and/or onto.



    So far, I have come up with a contradiction that proves $T$ is not one to one:



    $T(-1, 1, -1) = (-1, -1, 1)$
    and, $T(1, -1, 1) = (-1, -1, 1)$. However, I can't figure out whether or not $T$ is onto... I think it is, but I don't know the reaosn why or how to prove it.



    Can I say:



    Taking arbitrary elements a,b,c $in mathbb R$, $T(a,b,c) = (ab, bc, ac)$, and therefore any element $x,y,z in mathbb R$ can be generated by $x=ab$, $y = bc$, and $z=ac$?






    calculus functions linear-transformations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Given $T(x,y,z) = (xy, yz, xz)$, determine whether $T$ is one to one and/or onto.



      So far, I have come up with a contradiction that proves $T$ is not one to one:



      $T(-1, 1, -1) = (-1, -1, 1)$
      and, $T(1, -1, 1) = (-1, -1, 1)$. However, I can't figure out whether or not $T$ is onto... I think it is, but I don't know the reaosn why or how to prove it.



      Can I say:



      Taking arbitrary elements a,b,c $in mathbb R$, $T(a,b,c) = (ab, bc, ac)$, and therefore any element $x,y,z in mathbb R$ can be generated by $x=ab$, $y = bc$, and $z=ac$?






      calculus functions linear-transformations







      share|cite|improve this question









      $endgroup$




      Given $T(x,y,z) = (xy, yz, xz)$, determine whether $T$ is one to one and/or onto.



      So far, I have come up with a contradiction that proves $T$ is not one to one:



      $T(-1, 1, -1) = (-1, -1, 1)$
      and, $T(1, -1, 1) = (-1, -1, 1)$. However, I can't figure out whether or not $T$ is onto... I think it is, but I don't know the reaosn why or how to prove it.



      Can I say:



      Taking arbitrary elements a,b,c $in mathbb R$, $T(a,b,c) = (ab, bc, ac)$, and therefore any element $x,y,z in mathbb R$ can be generated by $x=ab$, $y = bc$, and $z=ac$?






      calculus functions linear-transformations




      calculus functions linear-transformations






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 7 '18 at 1:21









      JaigusJaigus

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          $begingroup$

          Assuming both the domain and codomain are $mathbb{R}^3$.



          Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.



          Suppose $$-1 =ab, -1=bc, -1=ac$$



          then we have $a = -frac{1}b, c= -frac{1}{b}$



          and we have $ac= frac1{b^2}>0$ which contracts $ac=-1<0$.






          share|cite|improve this answer









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            $begingroup$

            Assuming both the domain and codomain are $mathbb{R}^3$.



            Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.



            Suppose $$-1 =ab, -1=bc, -1=ac$$



            then we have $a = -frac{1}b, c= -frac{1}{b}$



            and we have $ac= frac1{b^2}>0$ which contracts $ac=-1<0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Assuming both the domain and codomain are $mathbb{R}^3$.



              Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.



              Suppose $$-1 =ab, -1=bc, -1=ac$$



              then we have $a = -frac{1}b, c= -frac{1}{b}$



              and we have $ac= frac1{b^2}>0$ which contracts $ac=-1<0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Assuming both the domain and codomain are $mathbb{R}^3$.



                Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.



                Suppose $$-1 =ab, -1=bc, -1=ac$$



                then we have $a = -frac{1}b, c= -frac{1}{b}$



                and we have $ac= frac1{b^2}>0$ which contracts $ac=-1<0$.






                share|cite|improve this answer









                $endgroup$



                Assuming both the domain and codomain are $mathbb{R}^3$.



                Suppose on the contrary that it is onto, then $(-1,-1,-1)$ has a preimage.



                Suppose $$-1 =ab, -1=bc, -1=ac$$



                then we have $a = -frac{1}b, c= -frac{1}{b}$



                and we have $ac= frac1{b^2}>0$ which contracts $ac=-1<0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 1:28









                Siong Thye GohSiong Thye Goh

                100k1466117




                100k1466117






























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