Question on judging a regular surface in differential geometry
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This is the proposition and proof saying that if we know that S is a regular surface and x is a candidate of parametrization which satisfies 1)differentiable 2)surjective differential map 3)continuous bijective then, x has a continuous inverse thus resulting that x is a real parametrization
However, in the proof, I cannot see that where I used the condition that S is a regular surface
Please let me know where the condition S is a regular surface used and is there any counter-example that x does not have a continuous inverse when S is not a regular surface?
condition 1 in the proposition is that x is differentiable
condition 2 is that locally homeomorphism
condition 3 is that differential dx is one-to-one
differential-geometry surfaces parametrization
asked Nov 14 at 6:56
Jaeyoon Yoo
876
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up vote
1
down vote
favorite
This is the proposition and proof saying that if we know that S is a regular surface and x is a candidate of parametrization which satisfies 1)differentiable 2)surjective differential map 3)continuous bijective then, x has a continuous inverse thus resulting that x is a real parametrization
However, in the proof, I cannot see that where I used the condition that S is a regular surface
Please let me know where the condition S is a regular surface used and is there any counter-example that x does not have a continuous inverse when S is not a regular surface?
condition 1 in the proposition is that x is differentiable
condition 2 is that locally homeomorphism
condition 3 is that differential dx is one-to-one
differential-geometry surfaces parametrization
asked Nov 14 at 6:56
Jaeyoon Yoo
876
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is the proposition and proof saying that if we know that S is a regular surface and x is a candidate of parametrization which satisfies 1)differentiable 2)surjective differential map 3)continuous bijective then, x has a continuous inverse thus resulting that x is a real parametrization
However, in the proof, I cannot see that where I used the condition that S is a regular surface
Please let me know where the condition S is a regular surface used and is there any counter-example that x does not have a continuous inverse when S is not a regular surface?
condition 1 in the proposition is that x is differentiable
condition 2 is that locally homeomorphism
condition 3 is that differential dx is one-to-one
differential-geometry surfaces parametrization
asked Nov 14 at 6:56
Jaeyoon Yoo
876
This is the proposition and proof saying that if we know that S is a regular surface and x is a candidate of parametrization which satisfies 1)differentiable 2)surjective differential map 3)continuous bijective then, x has a continuous inverse thus resulting that x is a real parametrization
However, in the proof, I cannot see that where I used the condition that S is a regular surface
Please let me know where the condition S is a regular surface used and is there any counter-example that x does not have a continuous inverse when S is not a regular surface?
condition 1 in the proposition is that x is differentiable
condition 2 is that locally homeomorphism
condition 3 is that differential dx is one-to-one
differential-geometry surfaces parametrization
differential-geometry surfaces parametrization
asked Nov 14 at 6:56
Jaeyoon Yoo
876
asked Nov 14 at 6:56
Jaeyoon Yoo
876
edited 23 hours ago
asked Nov 14 at 6:56
Jaeyoon Yoo
876
asked Nov 14 at 6:56
Jaeyoon Yoo
876
asked Nov 14 at 6:56
Jaeyoon Yoo
876
876
add a comment |
add a comment |
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