# Showing that the image of an arbitrary patch is an open set

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In summary, the student is trying to figure out how to get from the fact that every point in a surface has a neighborhood in the surface to the claim that x(D) is open.
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.

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:D\rightarrow 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^{-1}y## and ##y^{-1}x## are differentiable mappings defined on open sets of ##R^2##.

Let x be an arbitrary patch. Since M is a surface, for every point p there is a proper patch y whose image contains p. Based on the first part of the problem, y takes open sets to open sets. Based on the corollary, ##x^{-1}y## is defined on an open set, the set of points whose image, under y, are in the image of x. So the overlap of the images of x and y is an open set. But we need to show that the image of x is an open set, right?

I couldn't find the book on line. So what is a "proper patch" as opposed to an "arbitrary patch"?

lavinia said:
I couldn't find the book on line. So what is a "proper patch" as opposed to an "arbitrary patch"?
A patch is defined as a ono-to-one regular (the lower star is 1-to-1) map. It is proper if the inverse map is continuous.

Apparently we don't need the corollary to show that the image of an arbitrary patch in the surface M is open. Since every point of M is has a neighborhood in the image of a proper patch, every point in the image of an arbitrary patch in the surface has a neighborhood in the surface. So the image of an arbitrary patch is open.

The next problem asks the student to prove that every patch ##x:D\rightarrow M## in a surface M is proper. It says to use the results of the previous exercise (the one in this thread) and note that ##(x^{-1}y)y^{-1}## is continuous and agrees with ##x^{-1}## on an open set in x(D). I can see from the corollary that ##x^{-1}y## is continuous, and if y is a proper patch then ##y^{-1}## is continuous. But how to show that ##(x^{-1}y)y^{-1}## agrees with ##x^{-1}## on an open set? And how does this imply that x has a continuous inverse throughout x(D)?

Perhaps the "open set in x(D)" is the overlap of the images of x and y. That is the image, under y, of the domain of ##x^{-1}y##, an open set by the corollary. So then we would have ##x^{-1}## continuous on an open set in x(D). How to get from there to saying that it is continuous throughout x(D)?

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"Since every point of M is has a neighborhood in the image of a proper patch, every point in the image of an arbitrary patch in the surface has a neighborhood in the surface." that sounds suspiciously like, "A implies B so B implies A".

Well yes my statement seems a little garbled...

Let p be a point in the image of an arbitrary patch ##x:D\rightarrow M## in the surface.
p is a point in the surface.
There is a proper patch y such that a neighborhood of p is in the surface, by the definition of a surface.
Every point in x(D) has a neighborhood in M.

How to get from there to saying that these points have neighborhoods in x(D)?

martinbn said:
A patch is defined as a ono-to-one regular (the lower star is 1-to-1) map. It is proper if the inverse map is continuous.
I am confused what it means that the inverse is continuous. Does it mean that inverse images of open sets are intersections of open sets in the ambient Euclidean space with the image of the patch?

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That doesn't sound like what it means for the inverse to be continuous. "A coordinate patch ##x:d\rightarrow E^3## is a one-to-one regular mapping of an open set D of ##E^2## into ##E^3##. Proper patches are those for which the inverse function ##x^{-1}x(D)\rightarrow D## is continuous. That means that the limit of ##x^{-1}(q)## as q approaches p in x(D) equals ##x^{-1}(p)##.

## 1. What is the definition of an open set?

An open set is a subset of a topological space that does not contain its boundary points. This means that for every point in the set, there exists a neighborhood that is completely contained within the set.

## 2. Can you explain the concept of an arbitrary patch?

An arbitrary patch is a subset of a topological space that is homeomorphic to an open set in Euclidean space. This means that the patch can be continuously deformed into an open set without tearing or gluing any points together.

## 3. How can one show that the image of an arbitrary patch is an open set?

To prove that the image of an arbitrary patch is an open set, one needs to show that for every point in the image, there exists a neighborhood that is completely contained within the image. This can be done by using the continuity of the function that maps the patch to its image and the fact that the patch is homeomorphic to an open set.

## 4. What is the significance of showing that the image of an arbitrary patch is an open set?

Showing that the image of an arbitrary patch is an open set is important because it helps to establish the properties of the topological space. It also helps to understand the behavior of the function that maps the patch to its image, and can be used in various mathematical and scientific applications.

## 5. Are there any other related concepts that are important to understand in order to show that the image of an arbitrary patch is an open set?

Yes, there are a few other important concepts that are related to this proof. These include the notions of continuity and homeomorphism, which are essential in understanding the behavior of functions and subsets in a topological space. It is also helpful to have a basic understanding of open and closed sets, as well as boundary points and neighborhoods.

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