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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?