Can you give me an example of constant onto function [closed]












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What I think is a function which is constant but onto should just have the single value in codomain which the function takes






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closed as off-topic by Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos Dec 11 '18 at 11:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos

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    $begingroup$
    yes, you are correct
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    – BigbearZzz
    Dec 11 '18 at 8:11
















0












$begingroup$


What I think is a function which is constant but onto should just have the single value in codomain which the function takes






functions







share|cite|improve this question









$endgroup$



closed as off-topic by Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos Dec 11 '18 at 11:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    yes, you are correct
    $endgroup$
    – BigbearZzz
    Dec 11 '18 at 8:11














0












0








0


0



$begingroup$


What I think is a function which is constant but onto should just have the single value in codomain which the function takes






functions







share|cite|improve this question









$endgroup$




What I think is a function which is constant but onto should just have the single value in codomain which the function takes






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 8:10









You_know_whoYou_know_who

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closed as off-topic by Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos Dec 11 '18 at 11:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos Dec 11 '18 at 11:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Saad, GNUSupporter 8964民主女神 地下教會, Shaun, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    yes, you are correct
    $endgroup$
    – BigbearZzz
    Dec 11 '18 at 8:11














  • 2




    $begingroup$
    yes, you are correct
    $endgroup$
    – BigbearZzz
    Dec 11 '18 at 8:11








2




2




$begingroup$
yes, you are correct
$endgroup$
– BigbearZzz
Dec 11 '18 at 8:11




$begingroup$
yes, you are correct
$endgroup$
– BigbearZzz
Dec 11 '18 at 8:11










2 Answers
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So the codomain must have only $1$ element: $f:Cto D$, where $D={x_0}$. $C$, the domain, on the other hand, can be any nonempty set. So $f(x)=x_0,,forall xin C$.






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    $begingroup$

    Let $A=mathbb{Z}$ and $B={2},$ and define $f:A to B$ by $f(x)=2$ for any integer $x.$






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    $endgroup$




















      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      So the codomain must have only $1$ element: $f:Cto D$, where $D={x_0}$. $C$, the domain, on the other hand, can be any nonempty set. So $f(x)=x_0,,forall xin C$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        So the codomain must have only $1$ element: $f:Cto D$, where $D={x_0}$. $C$, the domain, on the other hand, can be any nonempty set. So $f(x)=x_0,,forall xin C$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          So the codomain must have only $1$ element: $f:Cto D$, where $D={x_0}$. $C$, the domain, on the other hand, can be any nonempty set. So $f(x)=x_0,,forall xin C$.






          share|cite|improve this answer









          $endgroup$



          So the codomain must have only $1$ element: $f:Cto D$, where $D={x_0}$. $C$, the domain, on the other hand, can be any nonempty set. So $f(x)=x_0,,forall xin C$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 8:23









          Chris CusterChris Custer

          13k3827




          13k3827























              1












              $begingroup$

              Let $A=mathbb{Z}$ and $B={2},$ and define $f:A to B$ by $f(x)=2$ for any integer $x.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $A=mathbb{Z}$ and $B={2},$ and define $f:A to B$ by $f(x)=2$ for any integer $x.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $A=mathbb{Z}$ and $B={2},$ and define $f:A to B$ by $f(x)=2$ for any integer $x.$






                  share|cite|improve this answer









                  $endgroup$



                  Let $A=mathbb{Z}$ and $B={2},$ and define $f:A to B$ by $f(x)=2$ for any integer $x.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 8:17









                  coffeemathcoffeemath

                  2,8451415




                  2,8451415















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