Fixing an orientation for a connected smooth surface

[Unconscious]

I am studying on Zorich, Mathematical Analysis II, 1st ed. pag. 174-175.
After having properly explained how orientations (equivalence classes) are defined for smooth k-dimensional surfaces in $\mathbb {R} ^ n$ that can be described with a single map, move on to the more general case by defining the meanings of:
1. consistent charts (two charts are consistent if their domains of action have empty intersection or if in the non-empty intersection the mutual transitions between one chart to the other have positive Jacobian),
2. orienting atlases (an orienting atlas is an atlas in which all charts are pairwise consistent),
3. equivalence classes for orienting atlases (possible orientations of the surface - a surface is oriented when an equivalence class in the set of all orienting atlases is fixed, that is, two atlases are equivalent if their union is an orienting atlas).

Having done this, he states without proof that a connected smooth k-dimensional surface can only have two possible orientations.
From this statement he immediately deduces that in order to fix an orientation on a surface of this type it is not necessary to exhibit an entire atlas of consistent charts, but it is sufficient to exhibit a single chart.

I was trying to prove why, but I can't.
I assumed, by absurdity, that I had two atlases of different orientation, made of pairwise consistent charts, containing a common chart $\varphi_1$:

$A_1=\{\varphi_1,\varphi_2,...,\varphi_m,...\}$
$A_2=\{\varphi_1,\phi_2,...,\phi_m,...\}$

but from here I can't get to any absurdity. Can anyone help me please?

 

[fresh_42]

I am studying on Zorich, Mathematical Analysis II, 1st ed. pag. 174-175.
After having properly explained how orientations (equivalence classes) are defined for smooth k-dimensional surfaces in $\mathbb {R} ^ n$ that can be described with a single map, move on to the more general case by defining the meanings of:
1. consistent charts (two charts are consistent if their domains of action have empty intersection or if in the non-empty intersection the mutual transitions between one chart to the other have positive Jacobian),
2. orienting atlases (an orienting atlas is an atlas in which all charts are pairwise consistent),
3. equivalence classes for orienting atlases (possible orientations of the surface - a surface is oriented when an equivalence class in the set of all orienting atlases is fixed, that is, two atlases are equivalent if their union is an orienting atlas).

Having done this, he states without proof that a connected smooth k-dimensional surface can only have two possible orientations.
From this statement he immediately deduces that in order to fix an orientation on a surface of this type it is not necessary to exhibit an entire atlas of consistent charts, but it is sufficient to exhibit a single chart.

I was trying to prove why, but I can't.
I assumed, by absurdity, that I had two atlases of different orientation, made of pairwise consistent charts, containing a common chart $\varphi_1$:

$A_1=\{\varphi_1,\varphi_2,...,\varphi_m,...\}$
$A_2=\{\varphi_1,\phi_2,...,\phi_m,...\}$

but from here I can't get to any absurdity. Can anyone help me please?

What do you want to prove, the Lemma that there are only two equivalence classes, or that it is sufficient to consider the first chart of an oriented atlas to determine the orientation?

It would be convenient to consider the problem in the language of differential forms to prove the latter. Do you have a definition of oriented manifolds in terms of differential forms, or just that the charts are equally oriented as mentioned above? And do we have a compact, connected manifold?

The idea is in both cases that we consider a path of charts from $x_0$ to $x$ (existence by connectivity, finiteness by compactness). Then all matrices of the basis changes from one chart to the next have a positive determinant. Multiplying them along the path doesn't change the sign, so the orientation of the first chart is all we need to know.

 

[Unconscious]

The day after posting the question I figured out how to show what I needed (that it is sufficient to consider the first chart of an oriented atlas to determine the orientation), and that's just the way you told me.
I also understood that, equivalent to displaying a single chart to fix the orientation of a connected smooth manifold, it is also possible to fix a frame at any point of that manifold.
But now, I'd like to go one step further and understand why, in addition to everything I've said above, my book also adds:

My doubt is: if every time I fix a normal $\mathbf {n}$, I force the other $n-1$ vectors $(\xi_1, ..., \xi_{n-1})$ (generators of the tangent space) to be such that the frame $F (\mathbf {n}) = (\mathbf {n}, \xi_1, ..., \xi_ {n-1})$ is in the same equivalence class as the frame $F_0 = (\mathbf {e} _1, ..., \mathbf {e} _ {n})$ (i.e., $\det M_ {F_0 \to F (\mathbf {n})}> 0$, with $M_ {F_0 \to F (\mathbf {n})}$ frame transition matrix), then if I take any two normals $\mathbf {n} _1$ and $\mathbf {n} _2$, I will always have that:

$\det M_{F(\mathbf{n}_2)\to F(\mathbf{n}_2)}=\det M_{F(\mathbf{n}_2)\to F_0}\cdot \det M_{F_0\to F(\mathbf{n}_1)}=\frac{\det M_{F_0\to F(\mathbf{n}_1)}}{\det M_{F_0\to F(\mathbf{n}_2)}}>0$

by construction, that is, I will always fix only one of the possible orientations of $S$.

 

[fresh_42]

I'm not quite sure where your question is. If we have an embedded surface, then there are two possibilities for the normal vector: outward bound, which is the standard choice, or inward bound, right hand rule versus left hand rule.

 

[Unconscious]

My question concerns the procedure that Zorich described in the annex to the previous message. I said that in my opinion that procedure (followed to the letter as described) produces frames always belonging to the same equivalence class, which instead one wants to avoid (I guess). I think instead that one wants to find a procedure such that, fixed a normal, a frame comes out in an equivalence class, while changing towards the normal a frame in the opposite equivalence class comes out.