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

• I
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?

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