Why is $\mathbb{R}P^n=D^n/\partial D^n$?

Question

This is just a question about understanding lines from textbook (A. Hatcher, Algebraic Topology):

We can restrict to vectors of length 1, so $\mathbb{R}$$P^n$ is also the quotient space $S^n/(v \thicksim -v)$, the sphere with antipodal points identified. This is equivalent to saying that $\mathbb{R}$$P^n$ is the quotient space of a hemisphere $D^n$ with antipodal points of $\partial D^n$ identified.

Here $\mathbb{R}$$P^n$ is the $\textbf{Real prjective n-space}$ defined to be the space of all lines through the origin in $\mathbb{R}^{n+1}$. I can understand that $\mathbb{R}P^n=S^n/(v \thicksim -v)$, but why is $\mathbb{R}P^n=D^n/\partial D^n$? If all of the points on the boundary of the disk are glued, what about the vectors inside the disk with different length?

Thank you!

Answer 1

Note that the hemisphere $D^n$ is "half of" $S^n$, namely $S^n=D^n\cup (-D^n)$ where $D^n\cap (-D^n)=\partial D^n$. Thus by using $D^n$ instead of $S^n$ we need not perform antipodal identification on most of the points of $D^n$, only on its boundary. (The result is not the same as $D^n/\partial D^n$ though!)

Answer 2

Vector length is irrelevant in topology. Anyway, it says that $D^n$ is a hemisphere, so all vectors do have the same length.

If $D^n$ is taken to be the Northern hemisphere, then the Southern hemisphere has been quotiented away, and the equator is $\partial D^n$.