Why do we ask for X to be open when defining a regular surface?
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Im reading do Carmo's geometry of curves and surfaces, which defines a regular surface as being (locally) the image of a function $X:R^2 to R^3$ satisfying:
1) $X$ is differentiable
2) $X$ is a homeomorphism
3) $dX$ is one-to-one
However, it seems to me that if a function has the listed properties except, perhaps, being open, we can still deduce openness from the remaining properties.
It is indeed proved in the book that, if one already knows a set $S$ to be a regular surface, then we need not ask for a parametrization $X$ to be a homeomorphism, but rather just one to one. The proof follows from the fact that regular surfaces are locally graphs.
But it can be shown, as i understand, that for a function satisfying conditions 1 and 2, its image set $S$ is also locally a graph.
If, in addition, we have said function be one to one, wouldn't that make $S$ a regular surface?
differential-geometry
asked Dec 16 '18 at 14:17
MaclioMaclio
314
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add a comment |
$begingroup$
Im reading do Carmo's geometry of curves and surfaces, which defines a regular surface as being (locally) the image of a function $X:R^2 to R^3$ satisfying:
1) $X$ is differentiable
2) $X$ is a homeomorphism
3) $dX$ is one-to-one
However, it seems to me that if a function has the listed properties except, perhaps, being open, we can still deduce openness from the remaining properties.
It is indeed proved in the book that, if one already knows a set $S$ to be a regular surface, then we need not ask for a parametrization $X$ to be a homeomorphism, but rather just one to one. The proof follows from the fact that regular surfaces are locally graphs.
But it can be shown, as i understand, that for a function satisfying conditions 1 and 2, its image set $S$ is also locally a graph.
If, in addition, we have said function be one to one, wouldn't that make $S$ a regular surface?
differential-geometry
asked Dec 16 '18 at 14:17
MaclioMaclio
314
$endgroup$
add a comment |
$begingroup$
Im reading do Carmo's geometry of curves and surfaces, which defines a regular surface as being (locally) the image of a function $X:R^2 to R^3$ satisfying:
1) $X$ is differentiable
2) $X$ is a homeomorphism
3) $dX$ is one-to-one
However, it seems to me that if a function has the listed properties except, perhaps, being open, we can still deduce openness from the remaining properties.
It is indeed proved in the book that, if one already knows a set $S$ to be a regular surface, then we need not ask for a parametrization $X$ to be a homeomorphism, but rather just one to one. The proof follows from the fact that regular surfaces are locally graphs.
But it can be shown, as i understand, that for a function satisfying conditions 1 and 2, its image set $S$ is also locally a graph.
If, in addition, we have said function be one to one, wouldn't that make $S$ a regular surface?
differential-geometry
asked Dec 16 '18 at 14:17
MaclioMaclio
314
$endgroup$
Im reading do Carmo's geometry of curves and surfaces, which defines a regular surface as being (locally) the image of a function $X:R^2 to R^3$ satisfying:
1) $X$ is differentiable
2) $X$ is a homeomorphism
3) $dX$ is one-to-one
However, it seems to me that if a function has the listed properties except, perhaps, being open, we can still deduce openness from the remaining properties.
It is indeed proved in the book that, if one already knows a set $S$ to be a regular surface, then we need not ask for a parametrization $X$ to be a homeomorphism, but rather just one to one. The proof follows from the fact that regular surfaces are locally graphs.
But it can be shown, as i understand, that for a function satisfying conditions 1 and 2, its image set $S$ is also locally a graph.
If, in addition, we have said function be one to one, wouldn't that make $S$ a regular surface?
differential-geometry
differential-geometry
asked Dec 16 '18 at 14:17
MaclioMaclio
314
asked Dec 16 '18 at 14:17
MaclioMaclio
314
edited Dec 16 '18 at 14:40
Maclio
asked Dec 16 '18 at 14:17
MaclioMaclio
314
asked Dec 16 '18 at 14:17
MaclioMaclio
314
asked Dec 16 '18 at 14:17
MaclioMaclio
314
314
add a comment |
add a comment |
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