Vtyjkyuk

This question is discussed in other threads but I don't follow the answers. Exchanging comments with the poster did not resolve my questions. The system took us to "chat", which also did not resolve my questions.

O'Neill's Elementary Differential Geometry, exercise 4.3.13 (first edition) poses the following problem:

"Prove that if $y:Erightarrow M$ is a proper patch, then y carries open sets to open sets in M. Deduce that if $x:Drightarrow M$ is an arbitrary patch, then the image x(D) is an open set in M. (Hint: To prove the latter assertion, use corollary 3.3)"

The thread proper patch says "The second part is clear if we prove the first part" with no explanation. It is not clear to me.

Corollary 3.3 is "If x and y are patches in a surface M whose images overlap, then the composite functions $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $E^2$.

I am doubtful of the first part of the problem but I am really asking about the second part. For the second part, I conclude that the intersection of the images of an arbitrary patch x and a proper patch y is an open set. But how does that imply that the image of x is an open set?

One of the other threads,every patch is proper says this is true "because of the definition of a surface." That shows that neighborhoods of a point in the image of x are contained in the surface but it doesn't seem to show that they are contained in the image of x. It also says "If a point $din D$ is part of a proper patch, we are done." I do not see why we are done. How does a point being part of a proper patch imply that the arbitrary patch is proper?

Part 1 - Proper patch carries open sets to open sets

Let a be an open set in E, and p be a point in a. p has a neighborhood in a. Consider the image of that neighborhood. It consists of more than a single point and has no holes in it, so $y(p)$ has a neighborhood in $y(a)$.

Part 2 - Image of an arbitrary patch is an open set

The hint is "If x and y are patches in a surface M in $R^3$ whose images overlap, then the composite functions $x^-1y$ and $y^-1x$ are differentiable mappings defined on open sets of $R^2$".

Let y be a proper patch that overlaps with an arbitrary patch x. Then $x^-1y$ is defined on an open set in E, and $x^-1$ is defined on an open set in M, namely O, the overlap of x and y. The overlap of x and y is an open set.

Let b be an open set in D. For every point $pin b$ there is a proper patch $y:Erightarrow y(E)$ such that $x(p)=y(q)$ and $x(b)cap y(E)$ is open. Since $x(p)in x(b)cap y(E)$ there is a neighborhood of x(p) that is contained in $x(b)cap y(E)$ and hence in $x(b)$. Thus x(b) is an open set.