Real Projective $n$ space $\mathbb{R}P^{n}$ In example 0.4 of Hatcher, he says that $\mathbb{R}P^{n}$ is just the quotient space of the sphere $S^{n}$ with antipodal points identified. He then says that this is equivalent to the quotient of a hemisphere $D^{n}$ with the antipodal points of the boundary identified. 
I don't understand why those spaces are equivalent. Could someone please explain? 
 A: Consider the quotient map $q : S^n \to \mathbb{R}P^n$. Consider also its restriction $q | D^n : D^n \to \mathbb{R}P^n$. Notice that for all $x,y \in D^n$, $q(x)=q(y)$ if and only if $x=y$ or $x,y$ are antipodal points on $\partial D^n$. Notice also that $q | D^n$ is a quotient map. Therefore, $\mathbb{R} P^n$ is homeomorphic to $D^n$ with antipodal points on its boundary identified.
A: So, a "point" in $\mathbb{RP}^n$ is secretly the same thing as a pair of antipodal points. But, if you look at two antipodal points in $S^n$, one of two things occurs:


*

*One of the points lies in the open hemisphere $\mathrm{Int}(D^n)$, and the other is in the opposite open hemisphere.

*Both points are on the equator $\partial D^n$.
So, every point in $\mathbb{RP}^n$ has a representative in $D^n$, and it has in fact only one such representative, except if it lies on the equator. So you get $\mathbb{RP}^n$ by considering $D^n$ and "correcting" the only injectivity default by collapsing the pairs of antipodal points in the equator.
A: To be very rigorous: view $D^n$ as an hemisphere included in $S^n$. The antipodal equivalence relation $\sim$ on $S^n$ restricts to the antipodal relation on $D^n$ (so we'll write $\sim$ for both). This means that the inclusion map $D^n \to S^n$ descends to a well-defined map $f:D^n / {\sim} \to S^n / {\sim}$ on the quotient spaces, and it is clear that $f$ is a bijection. By the definition of the quotient topologies on these spaces, $f$ is continuous. But, as a quotient of a compact space, $D^n/{\sim}$ is compact, and it is a good exercise to check by hand that $\mathbb{R}P^n = S^n/{\sim}$ is Hausdorff. Therefore $f$ is closed, continuous and bijective, hence a homeomorphism.
A: Every point on the sphere is mapped to the line in the direction of that point, hence antipodal points are identified as they map to the same line.
Similarly the hemisphere is also surjective on the set of lines via this identification. To make it a bijection (as before) one needs to identify only the antipodal points on the boundary, as the antipodal points to the internal points of the hemisphere lie outside the hemisphere, and so there (on the internal points) the map is already 1:1
