# Is there an intuitive way to see why $\mathbb{P}^2$ is non-orientable?

In the book of D. Chillingworth titled Differential Topology with a view to applications, at page 141, the author argues that

[...] $$\mathbb{P}^2$$ can be described equivalently as the set of all paris of antipodal points in $$S^2$$, [...] $$\mathbb{P}^2$$ is the quotient space $$S^2 / R$$ where $$R$$ is the equivalence relation "lying on the same line through the origin".

[...] The easiest way to see $$\mathbb{P}^2$$ is non-orientable is to think a coin heads uppermost near the north pole on $$S^2$$ being slid around the near the south pole and then being projected back to near the north pole directly by the antipodal map: it ends up tails uppermost.

However, I cannot understand the given intuitive argument about "projecting the coin by the antipodal map"; while we are sliding the coin towards the south pole, it is still head uppermost, and since the north and the south poles are the same points, being in the latter is the same as being in the former, so I don't see why the antipodal map needs to "flips" the coin.

Question:

What is the intuitive argument that the author is trying to convey ?

• The trick isn't whether the coin remains heads uppermost, but whether the coin rotating left is rotating right in its projection. You can think of a Möbius strip as a cylinder with antipodal points made equivalent - and, indeed, there is a subset of $\mathbb P^2$ which is a Möbius strip. – Thomas Andrews Apr 10 at 18:03
• @ThomasAndrews I'm quite familiar with those spaces, and I'm not asking about set theoretical explanation; just an intuitive one. – onurcanbektas Apr 10 at 18:26
• When you join the north pole to the south pole by a maximal semi-circle all the points that you are traveling belong to different equivalence classes in the quotient, except for the start and end points. Therefore, it is a loop in the quotient. Use an $\lfloor$ instead of a coin. Look at the sphere from above the north pole. When the $\lfloor$ ends at the south pole, what you see is $\rfloor$. This means that in the quotient you can start with an oriented frame $\lfloor$, do a loop and end up with the opposite orientation $\rfloor$. – user647486 Apr 10 at 18:33
• @user647486 Wow, that makes sense; thanks a lot. – onurcanbektas Apr 10 at 18:35