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

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## Main Question or Discussion Point

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?

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

martinbn
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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mathwonk
Homework Helper
"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)?

lavinia
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 $x^{-1}(p)$.