Is $f(x)=x+sin x$ a homeomorphic function?











up vote
0
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I want to determine $f(x) = x+sin x$ is homeomorphic or not on $mathbb{R}$?



A bijective continuous function is homeomorphic if its inverse is also continuous.
I know that $f$ is bijective. Also $f$ is continuous being the sum of two continuous functions.



How to look for the continuity of $f^{-1}$.






general-topology continuity inverse-function







share|cite|improve this question
























  • Which properties?
    – José Carlos Santos
    Nov 22 at 18:32










  • continuity of both sides.
    – Mittal G
    Nov 22 at 18:34






  • 1




    The statement follows from the inverse function theorem.
    – freakish
    Nov 22 at 18:38












  • @ José Carlos Santos see the edited part.
    – Mittal G
    Nov 22 at 18:38






  • 1




    @freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
    – Federico
    Nov 22 at 18:42















up vote
0
down vote

favorite
1












I want to determine $f(x) = x+sin x$ is homeomorphic or not on $mathbb{R}$?



A bijective continuous function is homeomorphic if its inverse is also continuous.
I know that $f$ is bijective. Also $f$ is continuous being the sum of two continuous functions.



How to look for the continuity of $f^{-1}$.






general-topology continuity inverse-function







share|cite|improve this question
























  • Which properties?
    – José Carlos Santos
    Nov 22 at 18:32










  • continuity of both sides.
    – Mittal G
    Nov 22 at 18:34






  • 1




    The statement follows from the inverse function theorem.
    – freakish
    Nov 22 at 18:38












  • @ José Carlos Santos see the edited part.
    – Mittal G
    Nov 22 at 18:38






  • 1




    @freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
    – Federico
    Nov 22 at 18:42













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I want to determine $f(x) = x+sin x$ is homeomorphic or not on $mathbb{R}$?



A bijective continuous function is homeomorphic if its inverse is also continuous.
I know that $f$ is bijective. Also $f$ is continuous being the sum of two continuous functions.



How to look for the continuity of $f^{-1}$.






general-topology continuity inverse-function







share|cite|improve this question















I want to determine $f(x) = x+sin x$ is homeomorphic or not on $mathbb{R}$?



A bijective continuous function is homeomorphic if its inverse is also continuous.
I know that $f$ is bijective. Also $f$ is continuous being the sum of two continuous functions.



How to look for the continuity of $f^{-1}$.






general-topology continuity inverse-function




general-topology continuity inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 18:44









egreg

176k1384198




176k1384198










asked Nov 22 at 18:26









Mittal G

1,182515




1,182515












  • Which properties?
    – José Carlos Santos
    Nov 22 at 18:32










  • continuity of both sides.
    – Mittal G
    Nov 22 at 18:34






  • 1




    The statement follows from the inverse function theorem.
    – freakish
    Nov 22 at 18:38












  • @ José Carlos Santos see the edited part.
    – Mittal G
    Nov 22 at 18:38






  • 1




    @freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
    – Federico
    Nov 22 at 18:42


















  • Which properties?
    – José Carlos Santos
    Nov 22 at 18:32










  • continuity of both sides.
    – Mittal G
    Nov 22 at 18:34






  • 1




    The statement follows from the inverse function theorem.
    – freakish
    Nov 22 at 18:38












  • @ José Carlos Santos see the edited part.
    – Mittal G
    Nov 22 at 18:38






  • 1




    @freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
    – Federico
    Nov 22 at 18:42
















Which properties?
– José Carlos Santos
Nov 22 at 18:32




Which properties?
– José Carlos Santos
Nov 22 at 18:32












continuity of both sides.
– Mittal G
Nov 22 at 18:34




continuity of both sides.
– Mittal G
Nov 22 at 18:34




1




1




The statement follows from the inverse function theorem.
– freakish
Nov 22 at 18:38






The statement follows from the inverse function theorem.
– freakish
Nov 22 at 18:38














@ José Carlos Santos see the edited part.
– Mittal G
Nov 22 at 18:38




@ José Carlos Santos see the edited part.
– Mittal G
Nov 22 at 18:38




1




1




@freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
– Federico
Nov 22 at 18:42




@freakish Not quite: the derivative vanishes at $x=(2k+1)pi$, $kinmathbb Z$.
– Federico
Nov 22 at 18:42










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










The function $f$ is continuous and strictly increasing, because its derivative is $ge0$ and is positive on the intervals $(pi+2kpi,pi+2(k+1)pi)$.



This its inverse function exists and is strictly increasing as well and defined over $mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because
$$
lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}
qquad
lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}
$$
Suppose that at some point $c$ the two limits are different, say
$$
lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}=a
qquad
lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}=b
$$
with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.






share|cite|improve this answer




























    up vote
    1
    down vote













    Once you know that $f$ is strictly increasing, just use this.



    If you want to know some regularity of $f^{-1}$, continue reading.



    Let $N={(2k+1)pi:kinmathbb Z}$.



    In $mathbb Rsetminus N$ you have $f'neq 0$, so $f$ is actually a diffeomorphism.
    Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $pi$. By Taylor expansion, we have that for $epsilon>0$ small enough
    $$
    |f(x)-f(pi)| = |x+sin(x)-pi| geq left|frac{(x-pi)^3}{12}right|
    qquad forall xin(pi-epsilon,pi+epsilon)
    $$
    so the inverse $f^{-1}$ is $tfrac13$-Holder near $pi$.






    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
      1
      down vote



      accepted










      The function $f$ is continuous and strictly increasing, because its derivative is $ge0$ and is positive on the intervals $(pi+2kpi,pi+2(k+1)pi)$.



      This its inverse function exists and is strictly increasing as well and defined over $mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because
      $$
      lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}
      qquad
      lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}
      $$
      Suppose that at some point $c$ the two limits are different, say
      $$
      lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}=a
      qquad
      lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}=b
      $$
      with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        The function $f$ is continuous and strictly increasing, because its derivative is $ge0$ and is positive on the intervals $(pi+2kpi,pi+2(k+1)pi)$.



        This its inverse function exists and is strictly increasing as well and defined over $mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because
        $$
        lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}
        qquad
        lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}
        $$
        Suppose that at some point $c$ the two limits are different, say
        $$
        lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}=a
        qquad
        lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}=b
        $$
        with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The function $f$ is continuous and strictly increasing, because its derivative is $ge0$ and is positive on the intervals $(pi+2kpi,pi+2(k+1)pi)$.



          This its inverse function exists and is strictly increasing as well and defined over $mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because
          $$
          lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}
          qquad
          lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}
          $$
          Suppose that at some point $c$ the two limits are different, say
          $$
          lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}=a
          qquad
          lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}=b
          $$
          with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.






          share|cite|improve this answer












          The function $f$ is continuous and strictly increasing, because its derivative is $ge0$ and is positive on the intervals $(pi+2kpi,pi+2(k+1)pi)$.



          This its inverse function exists and is strictly increasing as well and defined over $mathbb{R}$ because $f$ is neither upper nor lower bounded. In particular $f^{-1}$ has left and right limit at every point, because
          $$
          lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}
          qquad
          lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}
          $$
          Suppose that at some point $c$ the two limits are different, say
          $$
          lim_{xto c^-}f^{-1}(x)=sup{f(x):x<c}=a
          qquad
          lim_{xto c^+}f^{-1}(x)=inf{f(x):x>c}=b
          $$
          with $a<b$. Then $(a+b)/2$ doesn't belong to the codomain of $f^{-1}$, which is the domain of $f$. Contradiction.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 18:50









          egreg

          176k1384198




          176k1384198






















              up vote
              1
              down vote













              Once you know that $f$ is strictly increasing, just use this.



              If you want to know some regularity of $f^{-1}$, continue reading.



              Let $N={(2k+1)pi:kinmathbb Z}$.



              In $mathbb Rsetminus N$ you have $f'neq 0$, so $f$ is actually a diffeomorphism.
              Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $pi$. By Taylor expansion, we have that for $epsilon>0$ small enough
              $$
              |f(x)-f(pi)| = |x+sin(x)-pi| geq left|frac{(x-pi)^3}{12}right|
              qquad forall xin(pi-epsilon,pi+epsilon)
              $$
              so the inverse $f^{-1}$ is $tfrac13$-Holder near $pi$.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Once you know that $f$ is strictly increasing, just use this.



                If you want to know some regularity of $f^{-1}$, continue reading.



                Let $N={(2k+1)pi:kinmathbb Z}$.



                In $mathbb Rsetminus N$ you have $f'neq 0$, so $f$ is actually a diffeomorphism.
                Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $pi$. By Taylor expansion, we have that for $epsilon>0$ small enough
                $$
                |f(x)-f(pi)| = |x+sin(x)-pi| geq left|frac{(x-pi)^3}{12}right|
                qquad forall xin(pi-epsilon,pi+epsilon)
                $$
                so the inverse $f^{-1}$ is $tfrac13$-Holder near $pi$.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Once you know that $f$ is strictly increasing, just use this.



                  If you want to know some regularity of $f^{-1}$, continue reading.



                  Let $N={(2k+1)pi:kinmathbb Z}$.



                  In $mathbb Rsetminus N$ you have $f'neq 0$, so $f$ is actually a diffeomorphism.
                  Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $pi$. By Taylor expansion, we have that for $epsilon>0$ small enough
                  $$
                  |f(x)-f(pi)| = |x+sin(x)-pi| geq left|frac{(x-pi)^3}{12}right|
                  qquad forall xin(pi-epsilon,pi+epsilon)
                  $$
                  so the inverse $f^{-1}$ is $tfrac13$-Holder near $pi$.






                  share|cite|improve this answer












                  Once you know that $f$ is strictly increasing, just use this.



                  If you want to know some regularity of $f^{-1}$, continue reading.



                  Let $N={(2k+1)pi:kinmathbb Z}$.



                  In $mathbb Rsetminus N$ you have $f'neq 0$, so $f$ is actually a diffeomorphism.
                  Now we have to investigate what happens near $N$. By translation, it is sufficient to examine what happens around $pi$. By Taylor expansion, we have that for $epsilon>0$ small enough
                  $$
                  |f(x)-f(pi)| = |x+sin(x)-pi| geq left|frac{(x-pi)^3}{12}right|
                  qquad forall xin(pi-epsilon,pi+epsilon)
                  $$
                  so the inverse $f^{-1}$ is $tfrac13$-Holder near $pi$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 18:53









                  Federico

                  4,203512




                  4,203512






























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