Mixing
Jordan curve; is the inside mapped to the inside by a homeomorphism?
$begingroup$
Suppose that
$gamma:[0,1]to mathbb{R}^2$ is a closed simple curve;
$U$ is an open set of $mathbb{R}^2$ that contains both the image $|gamma|$ of $gamma$ and the inside $D$ of $gamma$;
$f:Uto V$ is a homeomorphism, where $V$ is an open set of $mathbb{R}^2$.
Is it true that the inside of the closed simple curve $fcircgamma $
is $f(D)$?
Intuitively, it seems true. But I can neither prove nor disprove it. Can anyone answer this question? Any help is appreciated.
general-topology
asked Jan 28 at 7:01
KenKen
186110
$endgroup$
add a comment |
$begingroup$
Suppose that
$gamma:[0,1]to mathbb{R}^2$ is a closed simple curve;
$U$ is an open set of $mathbb{R}^2$ that contains both the image $|gamma|$ of $gamma$ and the inside $D$ of $gamma$;
$f:Uto V$ is a homeomorphism, where $V$ is an open set of $mathbb{R}^2$.
Is it true that the inside of the closed simple curve $fcircgamma $
is $f(D)$?
Intuitively, it seems true. But I can neither prove nor disprove it. Can anyone answer this question? Any help is appreciated.
general-topology
asked Jan 28 at 7:01
KenKen
186110
$endgroup$
add a comment |
$begingroup$
Suppose that
$gamma:[0,1]to mathbb{R}^2$ is a closed simple curve;
$U$ is an open set of $mathbb{R}^2$ that contains both the image $|gamma|$ of $gamma$ and the inside $D$ of $gamma$;
$f:Uto V$ is a homeomorphism, where $V$ is an open set of $mathbb{R}^2$.
Is it true that the inside of the closed simple curve $fcircgamma $
is $f(D)$?
Intuitively, it seems true. But I can neither prove nor disprove it. Can anyone answer this question? Any help is appreciated.
general-topology
asked Jan 28 at 7:01
KenKen
186110
$endgroup$
Suppose that
$gamma:[0,1]to mathbb{R}^2$ is a closed simple curve;
$U$ is an open set of $mathbb{R}^2$ that contains both the image $|gamma|$ of $gamma$ and the inside $D$ of $gamma$;
$f:Uto V$ is a homeomorphism, where $V$ is an open set of $mathbb{R}^2$.
Is it true that the inside of the closed simple curve $fcircgamma $
is $f(D)$?
Intuitively, it seems true. But I can neither prove nor disprove it. Can anyone answer this question? Any help is appreciated.
general-topology
general-topology
asked Jan 28 at 7:01
KenKen
186110
asked Jan 28 at 7:01
KenKen
186110
asked Jan 28 at 7:01
KenKen
186110
asked Jan 28 at 7:01
KenKen
186110
asked Jan 28 at 7:01
KenKen
186110
186110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$Usetminus |gamma|$ may have several components $C$ (it's not given
that $U$ is connected), but only one with the property that $Ccup|gamma|$ is
compact, namely $C=D$. The image of $D$ under $f$ must have the same property
with respect to $V$, so $f(D)$ is the interior of $f(gamma)$.
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
$Usetminus |gamma|$ may have several components $C$ (it's not given
that $U$ is connected), but only one with the property that $Ccup|gamma|$ is
compact, namely $C=D$. The image of $D$ under $f$ must have the same property
with respect to $V$, so $f(D)$ is the interior of $f(gamma)$.
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
add a comment |
$begingroup$
$Usetminus |gamma|$ may have several components $C$ (it's not given
that $U$ is connected), but only one with the property that $Ccup|gamma|$ is
compact, namely $C=D$. The image of $D$ under $f$ must have the same property
with respect to $V$, so $f(D)$ is the interior of $f(gamma)$.
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
add a comment |
$begingroup$
$Usetminus |gamma|$ may have several components $C$ (it's not given
that $U$ is connected), but only one with the property that $Ccup|gamma|$ is
compact, namely $C=D$. The image of $D$ under $f$ must have the same property
with respect to $V$, so $f(D)$ is the interior of $f(gamma)$.
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$Usetminus |gamma|$ may have several components $C$ (it's not given
that $U$ is connected), but only one with the property that $Ccup|gamma|$ is
compact, namely $C=D$. The image of $D$ under $f$ must have the same property
with respect to $V$, so $f(D)$ is the interior of $f(gamma)$.
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Jan 28 at 7:16
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
add a comment |
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
$begingroup$
Thanks for your answer. How can we be sure that there is only one component of $U-|gamma|$ with the property stated in your answer?
$endgroup$
– Ken
Jan 28 at 8:13
add a comment |
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