If an immersion $X$ maps circles into planes then its image $X(mathbb{D})$ is homeomorphic to the cylinder.












4












$begingroup$


Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$
an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.



If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.



I do not know if all these assumptions are necessary, but that's what I have.



Can you help me prove that? I studied geometry and topology a long time ago.










share|cite|improve this question











$endgroup$












  • $begingroup$
    How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
    $endgroup$
    – ecrin
    Jan 12 at 20:29










  • $begingroup$
    $z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
    $endgroup$
    – Takashi
    Jan 13 at 0:44












  • $begingroup$
    If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
    $endgroup$
    – pre-kidney
    Jan 13 at 2:52










  • $begingroup$
    @pre-kidney. I corrected, thanks.
    $endgroup$
    – Takashi
    Jan 13 at 13:32
















4












$begingroup$


Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$
an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.



If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.



I do not know if all these assumptions are necessary, but that's what I have.



Can you help me prove that? I studied geometry and topology a long time ago.










share|cite|improve this question











$endgroup$












  • $begingroup$
    How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
    $endgroup$
    – ecrin
    Jan 12 at 20:29










  • $begingroup$
    $z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
    $endgroup$
    – Takashi
    Jan 13 at 0:44












  • $begingroup$
    If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
    $endgroup$
    – pre-kidney
    Jan 13 at 2:52










  • $begingroup$
    @pre-kidney. I corrected, thanks.
    $endgroup$
    – Takashi
    Jan 13 at 13:32














4












4








4


2



$begingroup$


Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$
an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.



If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.



I do not know if all these assumptions are necessary, but that's what I have.



Can you help me prove that? I studied geometry and topology a long time ago.










share|cite|improve this question











$endgroup$




Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$
an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.



If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.



I do not know if all these assumptions are necessary, but that's what I have.



Can you help me prove that? I studied geometry and topology a long time ago.







general-topology geometry analysis differential-geometry differential-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 18:22









Lord Shark the Unknown

104k1160132




104k1160132










asked Jan 7 at 23:32









TakashiTakashi

17917




17917












  • $begingroup$
    How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
    $endgroup$
    – ecrin
    Jan 12 at 20:29










  • $begingroup$
    $z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
    $endgroup$
    – Takashi
    Jan 13 at 0:44












  • $begingroup$
    If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
    $endgroup$
    – pre-kidney
    Jan 13 at 2:52










  • $begingroup$
    @pre-kidney. I corrected, thanks.
    $endgroup$
    – Takashi
    Jan 13 at 13:32


















  • $begingroup$
    How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
    $endgroup$
    – ecrin
    Jan 12 at 20:29










  • $begingroup$
    $z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
    $endgroup$
    – Takashi
    Jan 13 at 0:44












  • $begingroup$
    If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
    $endgroup$
    – pre-kidney
    Jan 13 at 2:52










  • $begingroup$
    @pre-kidney. I corrected, thanks.
    $endgroup$
    – Takashi
    Jan 13 at 13:32
















$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29




$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29












$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44






$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44














$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52




$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52












$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32




$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32










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