Problem
Prove that if $\mathbf{y}:E\to M$ is a proper patch, then $\mathbf{y}$ carries open sets in $E$ to open sets in $M$. Deduce that if $\mathbf{x}:D \to M$ is an arbitrary patch, then the image $\mathbf{x}(D)$ is an open set in $M$. (Hint: To prove the latter assertion, use Cor 3.3.)
Finally, prove that every patch $\mathbf{x}:D\to M$ in a surface $M$in $R^3$ is proper. (Hint: use the above to note that $(\mathbf{x}^{-1}\mathbf{y})\mathbf{y}^{-1}$ is continuous and agrees with $\mathbf{x}^{-1}$ on an open set in $\mathbf{x}(D)$.)
First I will write down the definitions of some of the terms above, as they are given in the text.
Definitions
A patch $\mathbf{x}:D\to R^3$ is a one-to-one regular mapping of an open set $D$ of $R^2$ into $R^3$.
A proper patch is a patch for which the inverse function $\mathbf{x}^{-1}:\mathbf{x}(D)\to D$ is continuous.
A surface in $R^3$is a subset $M$ of $R^3$ such that for each point $\mathbf{p}$ of $M$ there exists a proper patch in $M$ whose image contains a neighborhood of $\mathbf{p}$ in $M$.
For a function $F:R^n\to M$, each patch x in $M$ gives a coordinate expression $\mathbf{x}^{-1}(F)$ for $F$.
A function $F:R^n\to M$ is differentiable provided all its coordinate expressions are differentiable in the usual Euclidean sense.
3.3 Corollary
If x and y are patches in a surface $M$ in $R^3$ whose images overlap, then the composite functions $\mathbf{x}^{-1}\mathbf{y}$ and $\mathbf{y}^{-1}\mathbf{x}$ are differentiable mappings defined on open sets of $R^2$.
My Question
I don't know how to prove the first part.
Since y$^{-1}:\mathbf{y}(E)\to E$ is continuous, so by definition of continuity, given an open set $O$ in $E$,${\mathbf{y}^{-1}}^{-1}(O)=\mathbf{y}(O)$ is open in $\mathbf{y}(E)$, but how can we guarantee that this is open in $M$ when $\mathbf{y}(E)$ is not the whole of $M$?
Moreover, taking this as given, how can I use Cor 3.3 to prove that the image of an arbitrary patch is open in $M$? And finally, how can these be used to show that every patch in a surface is proper?
I think I'm confused with the topology in surfaces, and this is inhibiting my thinking. I would greatly appreciate it if anyone can write down a clear exposition of the above problem.
Best Answer
This is only a topology problem. Since that $M\subset \mathbb{R}^{3}$, $M$ have a relative topology induced by the topology on $\mathbb{R}^{3}$, $i.e.$, the open sets (or open neighborhood) $V$ on $M$ are open subsets on $\mathbb{R}^{3}$ intersection $M$ ($V=U\cap \mathbb{R}^{3}$). If $\chi: D\rightarrow M$ is a proper patch then this is a continuous function one-to-one with continuos inverse (topologically speaking), then $\chi$ is a homeomorphism between $D$ and $M$. Then $\chi(E)$ is open if $E\subset D$ is open. Another way is, W.L.G. $\chi(E)\subset \chi(D)\cap\rho(F)$ with $\rho:F\rightarrow M$ other proper patch. By your corolary $(\rho^{-1}\circ\chi)(E)$ is open since $\rho^{-1}\circ\chi$ is a difeomorphism on $\mathbb{R}^{3}$ and then $\rho\circ(\rho^{-1}\circ\chi)(E)$ is open on $M$, but $\rho\circ(\rho^{-1}\circ\chi)(E)=\chi(E)$ since $\rho$ and $\chi$ are one-to-one.