Proof that a proper patch carries open sets to open sets.
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O'Neill's Elementary Differential Geometry, problem 4.3.13 (Kindle edition), asks the student to show that the image of an open set, under a proper patch, is an open set.
Why is the image of a proper patch an open set discusses this question but does not specifically explain why. Here is my attempt at a solution. I do not know if it is complete as I have difficulty explaining the consequence of the fact that the mapping is continuous.
Let $phi$ be a proper patch and q be a point in $phi(E)$. Then $p=phi^-1(q)$ is in E. There is a neighborhood of p in E. All points sufficiently close to p map to points arbitrarily close to q. These points are in $phi(E)$. So there is a neighborhood of q that is in $phi(E)$ and $phi(E)$ is open. Specifically, $epsilon$ for the neighborhood of q is less than the maximum of $|phi(x)-q|$ for x in the neighborhood of p.
The problem goes on to ask the student to show that if $x:Drightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. There is a hint to use Corollary 4.3.3, which says "If x and y are patches in a surface M in $R^3$ whose images overlap, then the composite function $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $R^2$.
If x is an arbitrary patch and y is a proper patch, then it seems that the overlap of the images of x and y is an open set. But how to get from there to saying that the image of x is an open set?
differential-geometry
asked Sep 10 at 6:11
Gene Naden
637
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up vote
0
down vote
favorite
O'Neill's Elementary Differential Geometry, problem 4.3.13 (Kindle edition), asks the student to show that the image of an open set, under a proper patch, is an open set.
Why is the image of a proper patch an open set discusses this question but does not specifically explain why. Here is my attempt at a solution. I do not know if it is complete as I have difficulty explaining the consequence of the fact that the mapping is continuous.
Let $phi$ be a proper patch and q be a point in $phi(E)$. Then $p=phi^-1(q)$ is in E. There is a neighborhood of p in E. All points sufficiently close to p map to points arbitrarily close to q. These points are in $phi(E)$. So there is a neighborhood of q that is in $phi(E)$ and $phi(E)$ is open. Specifically, $epsilon$ for the neighborhood of q is less than the maximum of $|phi(x)-q|$ for x in the neighborhood of p.
The problem goes on to ask the student to show that if $x:Drightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. There is a hint to use Corollary 4.3.3, which says "If x and y are patches in a surface M in $R^3$ whose images overlap, then the composite function $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $R^2$.
If x is an arbitrary patch and y is a proper patch, then it seems that the overlap of the images of x and y is an open set. But how to get from there to saying that the image of x is an open set?
differential-geometry
asked Sep 10 at 6:11
Gene Naden
637
1
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
O'Neill's Elementary Differential Geometry, problem 4.3.13 (Kindle edition), asks the student to show that the image of an open set, under a proper patch, is an open set.
Why is the image of a proper patch an open set discusses this question but does not specifically explain why. Here is my attempt at a solution. I do not know if it is complete as I have difficulty explaining the consequence of the fact that the mapping is continuous.
Let $phi$ be a proper patch and q be a point in $phi(E)$. Then $p=phi^-1(q)$ is in E. There is a neighborhood of p in E. All points sufficiently close to p map to points arbitrarily close to q. These points are in $phi(E)$. So there is a neighborhood of q that is in $phi(E)$ and $phi(E)$ is open. Specifically, $epsilon$ for the neighborhood of q is less than the maximum of $|phi(x)-q|$ for x in the neighborhood of p.
The problem goes on to ask the student to show that if $x:Drightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. There is a hint to use Corollary 4.3.3, which says "If x and y are patches in a surface M in $R^3$ whose images overlap, then the composite function $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $R^2$.
If x is an arbitrary patch and y is a proper patch, then it seems that the overlap of the images of x and y is an open set. But how to get from there to saying that the image of x is an open set?
differential-geometry
asked Sep 10 at 6:11
Gene Naden
637
O'Neill's Elementary Differential Geometry, problem 4.3.13 (Kindle edition), asks the student to show that the image of an open set, under a proper patch, is an open set.
Why is the image of a proper patch an open set discusses this question but does not specifically explain why. Here is my attempt at a solution. I do not know if it is complete as I have difficulty explaining the consequence of the fact that the mapping is continuous.
Let $phi$ be a proper patch and q be a point in $phi(E)$. Then $p=phi^-1(q)$ is in E. There is a neighborhood of p in E. All points sufficiently close to p map to points arbitrarily close to q. These points are in $phi(E)$. So there is a neighborhood of q that is in $phi(E)$ and $phi(E)$ is open. Specifically, $epsilon$ for the neighborhood of q is less than the maximum of $|phi(x)-q|$ for x in the neighborhood of p.
The problem goes on to ask the student to show that if $x:Drightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. There is a hint to use Corollary 4.3.3, which says "If x and y are patches in a surface M in $R^3$ whose images overlap, then the composite function $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $R^2$.
If x is an arbitrary patch and y is a proper patch, then it seems that the overlap of the images of x and y is an open set. But how to get from there to saying that the image of x is an open set?
differential-geometry
differential-geometry
asked Sep 10 at 6:11
Gene Naden
637
asked Sep 10 at 6:11
Gene Naden
637
edited Sep 10 at 8:30
asked Sep 10 at 6:11
Gene Naden
637
asked Sep 10 at 6:11
Gene Naden
637
asked Sep 10 at 6:11
Gene Naden
637
637
1
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42
add a comment |Â
1
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42
1
1
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42
add a comment |Â
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1
Possible duplicate of Proving that every patch in a surface $M$ in $R^3$ is proper.
– Gene Naden
Sep 10 at 19:42