A question about injective continuous function and homeomorphism












1














Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?



$I=[0,1]$.$S^2$ is a 2-sphere.



I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...






general-topology







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    1














    Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?



    $I=[0,1]$.$S^2$ is a 2-sphere.



    I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...






    general-topology







    share|cite|improve this question

























      1












      1








      1







      Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?



      $I=[0,1]$.$S^2$ is a 2-sphere.



      I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...






      general-topology







      share|cite|improve this question













      Let $f:I rightarrow S^2$ be a continuous injective function.Does $I$ homeomorphic to $f(I)$?



      $I=[0,1]$.$S^2$ is a 2-sphere.



      I think it's right,but I don't know how to prove that $f(U)$ is a open subset in $f(I)$ if $U$ is a open subset in $I$.If it's wrong, I also cannot come up with counterexample...






      general-topology




      general-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 9:47









      MaxwellMaxwell

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      264






















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          $f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.






          share|cite|improve this answer

















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            Thank you very much!!
            – Maxwell
            Jan 4 at 10:02











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          1 Answer
          1






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          active

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          active

          oldest

          votes









          2














          $f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.






          share|cite|improve this answer

















          • 1




            Thank you very much!!
            – Maxwell
            Jan 4 at 10:02
















          2














          $f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.






          share|cite|improve this answer

















          • 1




            Thank you very much!!
            – Maxwell
            Jan 4 at 10:02














          2












          2








          2






          $f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.






          share|cite|improve this answer












          $f(I)$ is compact and any bijective continuous map between compact Hausdorff spaces is a homeomorphism. Instead of proving that $f(U)$ is open for any open set $U$ you can prove the equivalent statement that $f(C)$ is closed for any closed set $C$. Use the fact that $C$ is compact, so $f(C)$ is compact.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 9:49









          Kavi Rama MurthyKavi Rama Murthy

          51.8k32055




          51.8k32055








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            Thank you very much!!
            – Maxwell
            Jan 4 at 10:02














          • 1




            Thank you very much!!
            – Maxwell
            Jan 4 at 10:02








          1




          1




          Thank you very much!!
          – Maxwell
          Jan 4 at 10:02




          Thank you very much!!
          – Maxwell
          Jan 4 at 10:02


















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