- #1

Gene Naden

- 321

- 64

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?