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How we can prove that the North hemisphere is homeomorphic to the projective plane RP²?

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A hemisphere isn't homeomorphic to $\mathbb{R}P^2$, but a space homeomorphic to $\mathbb{R}P^2$ is obtained by identifying the antipodal points on the boundary of the hemisphere. – Brad Sep 28 '12 at 12:50
If you ask a question that's closely related to a question you asked previously, it makes sense to link to that other question; this provides context and may prevent unnecessary duplication of efforts. – joriki Sep 28 '12 at 12:52
up vote 2 down vote accepted

If you are thinking about hemispheres then the right way to look at $\Bbb{R}P^2$ like this is using $CW$ - complexes. We have a $CW$ - structure on $\Bbb{R}P^2$ (see Hatcher page 6) with

$$\Bbb{R}P^2 = \Bbb{R}P^1 \sqcup D^2/ (x \sim \varphi(x))$$

where $\varphi : S^1 \to \Bbb{R}P^1$ is the attaching map that so happens to be the quotient map that identifies antipodal points. In other words, you can think of $\Bbb{R}P^2$ as first you take a circle, and then you paste $D^2$ onto $S^1$ in the usual way but now the pasting is funny: Points on the boundary of $D^2$ have been identified with their antipodes. Now stretch your $D^2$ vertically and puff it up a little so it looks like a hemisphere.

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