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Guillemin and Pollack define an orientation on a $k$-dimensional manifold $X\subset \mathbb R^N$ with boundary is a smooth choice of orientations for all tangent spaces $T_x(X)$. (The smoothness conditions means this: around each $x\in X$ there must exists a local parametrization $h: U\to X$ such that $dh_u: \mathbb R^k\to T_{h(u)}(X)$ preserves orientation at each $u\in U\subset H^k=\{(x_1,\dots,x_k): x_k \ge 0\}$.

They then claim that the Moebius strip isn't orientable. "You are invited to make a paper model and "prove" pictorially that no smooth orientation of the Moebius strip exists. The difficulty is that if you walk around a transparent strip pitching pennies heads up, eventually you return to the starting point to find tails up!"

I have no idea how their "argument" with heads and tails is connected with the definition they gave, and what it means mathematically that "we walk around a strip".

Another (actually, I guess the same) argument I've seen is that one take a normal vector, "translates" it around the strip, and notes that the direction is changed. But again I don't understand what such argument contradicts to (and again what it means to translate the vector).

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  • $\begingroup$ They assume you can see through the paper and that the paper has no thickness. Their metaphor and intuition are not consistent. $\endgroup$ – The Count Mar 28 '18 at 0:29
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The pennies are meant as metaphors for those local parametrizations. You start with heads up, and continue to align the next with the last: orientations are preserved. So they are all have heads up. After the whole trip, you see the first one and it is tails up. You can't align the last one with the first one. That is the contradiction.

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