Embedding Euclidean Space Into Real Projective Space

Question

I'm struggling with what I think should be a pretty straightforward proof.

Let $g:\mathbb{R}^n\rightarrow\mathbb{R}P^n$ be the map defined by $g=p\circ f$ where $p$ is the quotient map $p:\mathbb{R}^{n+1}\rightarrow\mathbb{R}P^n$ and $f(x_1,x_2,\cdots,x_n)=(1,x_1,x_2,\cdots,x_n)$. I'm trying to show that this map is an embedding.

I thought I had proved it, but I noticed that in trying to prove $g$ is an open map into its image, I assumed that the restriction of $p$ to $f(\mathbb{R}^n$) was a quotient map (which I cannot do since $f(\mathbb{R}^n)$ is not saturated). I'm having trouble figuring out how to get around this.

Answer 1

Hint: Let $U\subset\mathbb{R}^{n+1}\setminus\{0\}$ be the set of points with nonzero first coordinate. Then $U$ is a saturated open set, so $p(U)$ is open and $p$ restricts to a quotient map $U\to p(U)$. Now can you show that $p|_U\circ f:\mathbb{R}^n\to p(U)$ is a homeomorphism by explicitly describing its inverse in terms of a map from $U$?

More details are hidden below.

Consider the map $h:U\to \mathbb{R}^n$ given by $h(x_0,x_1,\dots,x_n)=(x_1/x_0,\dots,x_n/x_0)$. Then $h$ is constant on equivalence classes and so induces a continuous map $p(U)\to \mathbb{R}^n$, which is inverse to $p|_U\circ f$.