Boundary of a surface-with-boundary
I'm reading "Advanced calculus: a differential forms approach".
Author defines $S$, a compact, oriented, differentiable surface-with-boundary in $xyz$-space by describing a finite number of oriented charts $F_i$, of the $uv$-plane to $xyz$-plane. In particular,
(a) A finite number of differentiable maps ("charts") $F_1, F_2, ..., F_N$ of the $uv$-plane to the $xyz$-space are given. Each $F_i$ is one-to-one and non-singular of rank 2 on the square $ \{|u|≤ 1, |v|≤1\}.$
(b) For each $F_i$ there is specified a closed rectactangle in $ \{|u|≤ 1, |v|≤1\}$ , such that the image of a point of the square under $F_i$ is a point of $S$ if and only if the point lies in $R_i$.
(c) For each $P$ in $S$ there is an $i, 1≤ i ≤ N $, and an $\epsilon>0$, such that $P=F_i(u, v)$ where $|u|<1, |v|<1$, and such that a point $Q$ in the $xyz$-space which lies within $\epsilon$ of $P$ lies in $S$ if and only if it lies in $F_i(R_i)$
(d) the orientations of the charts agree.
It is then defined the boundary of $S$, $\partial S$, as
Those points of $S$ which are the image under some $F_i$ of a point inside $ \{|u|≤ 1, |v|≤1\}$ which lies on a boundary of $R_i$.
How is this possible? Why a point on a boundary of $R_i$, under $F_i$, is a point in $\partial S$?
Boundary of a sphere, according to this definition
To visualize the definition, I tried to apply it to a sphere, to see what would happen.
Let's consider S as the unit sphere centered at the origin, i.e.
$S=\{(x,y,z)∈\mathbb{R}^3∣x^2+y^2+z^2=1\}$
We can choose, using the streographic projections of the sphere minus the north (or south) pole, and by introducing a scale factor, $F_1, F_2$ so that they parametrize the square one-to-one into an entire hemisphere. Hence, $R_1=R_2=\{|u|≤ 1, |v|≤1\}.$ (Of course we have also to make sure that these charts are consistently oriented).
But then, by definition, the boundary of S, $\partial S$, is
$\bigcup\limits_{i=1}^{2} F_{i}(\partial R_i)$;
i.e. the equator of the sphere. But this is of course false, the (manifold's) boundary of a sphere is the empty set. What if we were to define the charts such that they parametrize all but a small area around the omitted pole? The boundary of the sphere would change (hence this definition is not well-defined); and then would the Stokes's Theorem hold?
Screenshots of the pages I'm talking about:
The answer is that the author managed to give a wrong definition of a surface with boundary. After a bit of thought, item (c) in the definition implies that each (closed rectangle) $R_i$ is contained in the interior of the square $[-1,1]^2$. From this, it follows (again using item (c)) that the images of the rectangles $R_i$ under the maps $F_i$ are pairwise disjoint in $S$. Thus, what the author of the book "defined" is not a surface with boundary but a surface with corners. Moreover, only few surfaces with corners satisfy this definition (only those which are diffeomorphic to the disjoint union of finitely many closed rectangles), for instance, $S^2$, $T^2$, etc., do not.
There are many sources where surfaces (and, more generally, manifolds) with boundary are defined properly. For instance, consider reading the book by Guillemin and Pollack "Differential Topology."