Sufficient condition for onto.

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I want to ask you for following statement




f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?




My attempt



For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
$B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto



First, I want to ask whether this proof is correct or not.



Second, Is there any counter example if X has only connect condition ?



Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$



Thank you!










share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    I want to ask you for following statement




    f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?




    My attempt



    For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
    $B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto



    First, I want to ask whether this proof is correct or not.



    Second, Is there any counter example if X has only connect condition ?



    Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$



    Thank you!










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I want to ask you for following statement




      f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?




      My attempt



      For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
      $B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto



      First, I want to ask whether this proof is correct or not.



      Second, Is there any counter example if X has only connect condition ?



      Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$



      Thank you!










      share|cite|improve this question













      I want to ask you for following statement




      f is continuous from X to Y, where X is compact Hausdorff, connected and Y is ordered set in order topology, then f is onto?




      My attempt



      For given $y in Y$ choose nbhd $N_y$. Since $f$ is continuous and $X$ is compact Hausdorff,
      $B=f^-1(bar N_y$) is compact in X , Where $bar N_y$ is closure of nbhd. Since $f(B)$ is compact on Y which contain $y$, it has extreme value. Now we can apply Intermediate theorem on $f(B)$ , so we can find $x^* in B$ such that $f(x*)=y$ . Thus $f$ is onto



      First, I want to ask whether this proof is correct or not.



      Second, Is there any counter example if X has only connect condition ?



      Third, Is there any other condition $f$ could be automatically onto for specific condition of $X$ and $Y$



      Thank you!







      general-topology






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      asked 1 hour ago









      fivestar

      31519




      31519




















          2 Answers
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          Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.



          A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).






          share|cite|improve this answer




















          • Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
            – fivestar
            1 hour ago

















          up vote
          2
          down vote













          Your proof is not correct as pointed out by Mees de Vries.



          However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.



            A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).






            share|cite|improve this answer




















            • Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
              – fivestar
              1 hour ago














            up vote
            3
            down vote













            Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.



            A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).






            share|cite|improve this answer




















            • Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
              – fivestar
              1 hour ago












            up vote
            3
            down vote










            up vote
            3
            down vote









            Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.



            A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).






            share|cite|improve this answer












            Your proof is not correct; you assume (incorrectly) that $B$ is non-empty.



            A very simple counterexample to your statement is the inclusion $[0, 1] to [0, 2]$ with the ordinary Euclidean/order topology (an even simpler example is the inclusion $0 to 0, 1$, assuming that $Y$ need not be connected).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Mees de Vries

            15k12450




            15k12450











            • Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
              – fivestar
              1 hour ago
















            • Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
              – fivestar
              1 hour ago















            Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
            – fivestar
            1 hour ago




            Your right , I made a mistake ! Thanks for feedback. and I under stand your example!
            – fivestar
            1 hour ago










            up vote
            2
            down vote













            Your proof is not correct as pointed out by Mees de Vries.



            However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.






            share|cite|improve this answer
























              up vote
              2
              down vote













              Your proof is not correct as pointed out by Mees de Vries.



              However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                Your proof is not correct as pointed out by Mees de Vries.



                However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.






                share|cite|improve this answer












                Your proof is not correct as pointed out by Mees de Vries.



                However, take any non-empty space $X$. Take any order topology on a space $Y$ with at least $2$ points, say $y$ and $y'$. Define $f:Xrightarrow Y$ by $f(x)=y$. Then, of course $f$ is not onto and any property(in particular connectedness, compactness or Hausdorfness) of the topological space $X$ does not make a difference in this regard.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                sqtrat

                1,888518




                1,888518



























                     

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