Let $S$ be the unit circle in the complex plane and $f :[0,2]rightarrow S$ be defined as $f(t)=e^2pi i t$

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Let $S$ be the unit circle in the complex plane with the induced topology from $mathbbC$
and let $f :[0,2]rightarrow S$ be defined as $f(t)=e^2pi i t$



Then which of the following statements are true?



(a) $K$ is closed in $[0,2] $ implies $ f(K) $ is closed in $S$.



(b) $f(X)$ is closed in $S$ implies $ X$ is closed in $[0,2]$.



(c) $U$ is open in $[0,2] $ implies $ f(U) $ is open in $S$.



(d) $f(X)$ is open in $S$ implies $X$ is open in $[0,2]$.



My efforts:



As $f$ is a continuous map the inverse image of open/closed is open/closed so (d) and (b) are true.



What about other cases? I think they don't need to be true.




complex-analysis




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  • Actually, b) and d) are false and $a),c) are true!
    – Kavi Rama Murthy
    Aug 29 at 5:41











  • Can you give me a hint?
    – StammeringMathematician
    Aug 29 at 5:42






  • 1




    Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 29 at 5:46










  • Sorry, there was an error in my first comment. c) is also false. See my answer for details.
    – Kavi Rama Murthy
    Aug 29 at 5:49














up vote
0
down vote

favorite
1












Let $S$ be the unit circle in the complex plane with the induced topology from $mathbbC$
and let $f :[0,2]rightarrow S$ be defined as $f(t)=e^2pi i t$



Then which of the following statements are true?



(a) $K$ is closed in $[0,2] $ implies $ f(K) $ is closed in $S$.



(b) $f(X)$ is closed in $S$ implies $ X$ is closed in $[0,2]$.



(c) $U$ is open in $[0,2] $ implies $ f(U) $ is open in $S$.



(d) $f(X)$ is open in $S$ implies $X$ is open in $[0,2]$.



My efforts:



As $f$ is a continuous map the inverse image of open/closed is open/closed so (d) and (b) are true.



What about other cases? I think they don't need to be true.




complex-analysis




share|cite|improve this question






















  • Actually, b) and d) are false and $a),c) are true!
    – Kavi Rama Murthy
    Aug 29 at 5:41











  • Can you give me a hint?
    – StammeringMathematician
    Aug 29 at 5:42






  • 1




    Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 29 at 5:46










  • Sorry, there was an error in my first comment. c) is also false. See my answer for details.
    – Kavi Rama Murthy
    Aug 29 at 5:49












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Let $S$ be the unit circle in the complex plane with the induced topology from $mathbbC$
and let $f :[0,2]rightarrow S$ be defined as $f(t)=e^2pi i t$



Then which of the following statements are true?



(a) $K$ is closed in $[0,2] $ implies $ f(K) $ is closed in $S$.



(b) $f(X)$ is closed in $S$ implies $ X$ is closed in $[0,2]$.



(c) $U$ is open in $[0,2] $ implies $ f(U) $ is open in $S$.



(d) $f(X)$ is open in $S$ implies $X$ is open in $[0,2]$.



My efforts:



As $f$ is a continuous map the inverse image of open/closed is open/closed so (d) and (b) are true.



What about other cases? I think they don't need to be true.




complex-analysis




share|cite|improve this question














Let $S$ be the unit circle in the complex plane with the induced topology from $mathbbC$
and let $f :[0,2]rightarrow S$ be defined as $f(t)=e^2pi i t$



Then which of the following statements are true?



(a) $K$ is closed in $[0,2] $ implies $ f(K) $ is closed in $S$.



(b) $f(X)$ is closed in $S$ implies $ X$ is closed in $[0,2]$.



(c) $U$ is open in $[0,2] $ implies $ f(U) $ is open in $S$.



(d) $f(X)$ is open in $S$ implies $X$ is open in $[0,2]$.



My efforts:



As $f$ is a continuous map the inverse image of open/closed is open/closed so (d) and (b) are true.



What about other cases? I think they don't need to be true.





complex-analysis





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edited Aug 29 at 6:05









dmtri

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asked Aug 29 at 5:33









StammeringMathematician

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  • Actually, b) and d) are false and $a),c) are true!
    – Kavi Rama Murthy
    Aug 29 at 5:41











  • Can you give me a hint?
    – StammeringMathematician
    Aug 29 at 5:42






  • 1




    Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 29 at 5:46










  • Sorry, there was an error in my first comment. c) is also false. See my answer for details.
    – Kavi Rama Murthy
    Aug 29 at 5:49
















  • Actually, b) and d) are false and $a),c) are true!
    – Kavi Rama Murthy
    Aug 29 at 5:41











  • Can you give me a hint?
    – StammeringMathematician
    Aug 29 at 5:42






  • 1




    Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
    – Ð°ÑÑ‚он вілла олоф мэллбэрг
    Aug 29 at 5:46










  • Sorry, there was an error in my first comment. c) is also false. See my answer for details.
    – Kavi Rama Murthy
    Aug 29 at 5:49















Actually, b) and d) are false and $a),c) are true!
– Kavi Rama Murthy
Aug 29 at 5:41





Actually, b) and d) are false and $a),c) are true!
– Kavi Rama Murthy
Aug 29 at 5:41













Can you give me a hint?
– StammeringMathematician
Aug 29 at 5:42




Can you give me a hint?
– StammeringMathematician
Aug 29 at 5:42




1




1




Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 29 at 5:46




Well, $f$ is a continuous map, but it is not bijective, so $f^-1 (f(X))$ is not equal to $X$ for all open sets $X$ i.e. the inverse image of the set $f(X)$ is not $X$. You need to be more careful while tackling $b$ and $d$.
– Ð°ÑÑ‚он вілла олоф мэллбэрг
Aug 29 at 5:46












Sorry, there was an error in my first comment. c) is also false. See my answer for details.
– Kavi Rama Murthy
Aug 29 at 5:49




Sorry, there was an error in my first comment. c) is also false. See my answer for details.
– Kavi Rama Murthy
Aug 29 at 5:49










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For a) note that $f$ maps compact sets to compact sets. So a) is true. For b) use the fact that $f([0,1))=S$ but $[0,1)$ is not closed. For c) use that fact that $[0,frac 1 2)$ is an open subsety of $[0,2]$ but $f([0,frac 1 2))$is not open. Finally, $f((0,1])=S$ so d) is false.






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    For a) note that $f$ maps compact sets to compact sets. So a) is true. For b) use the fact that $f([0,1))=S$ but $[0,1)$ is not closed. For c) use that fact that $[0,frac 1 2)$ is an open subsety of $[0,2]$ but $f([0,frac 1 2))$is not open. Finally, $f((0,1])=S$ so d) is false.






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      up vote
      2
      down vote



      accepted










      For a) note that $f$ maps compact sets to compact sets. So a) is true. For b) use the fact that $f([0,1))=S$ but $[0,1)$ is not closed. For c) use that fact that $[0,frac 1 2)$ is an open subsety of $[0,2]$ but $f([0,frac 1 2))$is not open. Finally, $f((0,1])=S$ so d) is false.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        For a) note that $f$ maps compact sets to compact sets. So a) is true. For b) use the fact that $f([0,1))=S$ but $[0,1)$ is not closed. For c) use that fact that $[0,frac 1 2)$ is an open subsety of $[0,2]$ but $f([0,frac 1 2))$is not open. Finally, $f((0,1])=S$ so d) is false.






        share|cite|improve this answer












        For a) note that $f$ maps compact sets to compact sets. So a) is true. For b) use the fact that $f([0,1))=S$ but $[0,1)$ is not closed. For c) use that fact that $[0,frac 1 2)$ is an open subsety of $[0,2]$ but $f([0,frac 1 2))$is not open. Finally, $f((0,1])=S$ so d) is false.







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        share|cite|improve this answer










        answered Aug 29 at 5:45









        Kavi Rama Murthy

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