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I saw this in Basic Topology by M.A.Armstrong. It gives three descriptions of real n-dimensional projective space $P^n$. Two of them are:

(a) Begin with the unit sphere $S^n$ in $E^{n+1}$ and identify its antipodal points.

(c) Begin with the unit ball $B^n$ and identify antipodal points of its boundary sphere.

I find it hard to imagine why these two descriptions lead to the same space.

Can you please help? Thank you.

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    $\begingroup$ (a) is the same as the upper hemisphere, where you identify antipodal points on the boundary $S^{n-1}$ - and that in turn is the same as (b) $\endgroup$ – user8268 Mar 29 '11 at 14:32
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    $\begingroup$ I would say (c) gives you n-1 dimensional projective space, while (a) gives you n dimensional. $\endgroup$ – GEdgar Mar 29 '11 at 14:42
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    $\begingroup$ @GEdgar , that's not true. In (c), only the antipodal points of the boundary sphere are identified. $\endgroup$ – Roun Mar 29 '11 at 14:59
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Start with (a). Given $S^n$, first think about all points not on the equator (here, if $S^n = \{(x_1,...,x_{n+1})|$ $x_1^2 + ... + x_{n+1}^2 = 1\}$, then the equator is all the points with, say, $x_{n+1} = 0$).

When we identify these particular points, every point has a unique representative in the open "northern" (i.e., $x_{n+1} > 0) hemisphere. We still need to make identifications on the equatorial boundary of the closed northern hemisphere.

Thus, we can obtain $\mathbb{R}P^n$ by taking just the northern hemisphere of a sphere and identitifying some more points on the equator. But the northern hemisphere is a (closed) n-ball, and the equator is the boundary of the n-ball. Finishing up the identificatin on $S^n$ is simply a matter of identifying antipodal points on the equator, but the equator is the boundary of the n-ball, so the two constructions give the same space.

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