Embedding Real Projective Space in Sphere 
Let $n > 1$ be an integer and set $N =\frac{n(n+1)
}{2}$. Let $F : R^n \setminus \{0\} → R^N$ be defined by
$(x_i)_i
\to (x_ix_j)_{i \leq j}$
, where the $x_i x_j$ are ordered lexicographically.


(a) Show that $F$ is an immersion.


(b) Show that $F(v) = F(w)$ if and only if $v = \pm w$. Show that $F$ induces an embedding
$f : RP^{n−1} \to S^{N−1}$.


(c) Show that $f$ cannot be surjective. Conclude that $f$ induces an embedding of $RP^{n−1}$
in $R^{N−1}$.

So I solved $(a)$ by considering the Jacobian matrix, and by showing that the rank of the Jacobian matrix is greater than or equal to $n$. This should be sufficent, Isn't it?
Regarding $(b)$, I was able to prove that $F(v)=F(w) \Rightarrow v=\pm w$. How do I show that $f(\mathbb{R}P^{n-1}) \subset S^{N-1}$. I think that since we are taking elements of $\mathbb{R}P^{n-1}$, then as per the definition of the map,for any $x=(x_1,x_2,x_3 \dots x_n)$ the norm of $f(x)$ is homogenous in $x_1,x_2 \dots x_n$ . Since I am working in $\mathbb{R}P^{n-1}$, I can choose $x$ such that $||f(x)=1||$ and that shows that the image lies in $S^{N-1}$? Is this correct? Also how do I show that the map $f$ is an immersion?
Also I have no idea with $(c)$. I request someone to verify $(a)$ and help me with $(b)$ and $(c)$
 A: (a) Your approach is correct. Note that the rank  of the Jacobian matrix cannot be greater than $n$ because you have only $n$ coordinates.
(b) The purpose is to show that $F$ induces an embedding $f : \mathbb RP^{n−1} \to S^{N−1}$. Since $F(v) = F(w)$ if and only if $v = \pm w$, one might naively think that $f([x]) = F(x)$ for $x \in S^{n-1}$ is the "correct" definition. Note that each equivalence class $\xi \in \mathbb RP^{n−1}$ can be represented by an up to sign unique $x \in S^{n-1}$. But the problem is that in general $F(x) \notin S^{N−1}$ if $x \in S^{n−1}$. As an example take $x = \frac{1}{\sqrt n}(1,\dots,1) \in S^{m−1}$. Then all $x_i x_j = \frac{1}{n}$ and $\sum_{i\le j}(x_i x_j)^2 = \frac{n(n+1)}{2}\frac{1}{n^2} = \frac{n+1}{2n} < 1$ for $n > 1$. We thus need another approach.
Let us first observe that $F(x) \ne 0$ for all $x \in \mathbb R^n \setminus \{0\}$. This is true because $x$ has a coordinate $x_i \ne 0$ and thus $F(x)$ has the coordinate $F_{ii}(x) \ne 0$. Therefore we should consider the map
$$G : R^n \setminus \{0\} \to \mathbb R^N, G(x) = \dfrac{F(x)}{\lVert F(x) \rVert} .\tag {1}$$
This is well-defined and smooth map. Since $G(x) \in S^{N-1}$ and $S^{N-1}$ is a smooth submanifold of $\mathbb R^N$ , it gives us a smooth map $\tilde G : R^n \setminus \{0\} \to S^{N-1}$ which clearly induces a continuous $f : \mathbb RP^{n-1} \to S^{N-1}$ (since $[v] = [w]$ means $v = \lambda w$ for some $\lambda \ne 0$ and thus $F(v) = \lambda^2 F(w)$ which implies $G(v) = G(w)$). The quotient map $p : R^n \setminus \{0\} \to \mathbb RP^{n-1}$ is a smooth submersion, hence $f$ is smooth. See for example Proposition 5.20 "(Passing Smoothly to the Quotient)" in John M. Lee's INTRODUCTION TO SMOOTH MANIFOLDS.
Let us prove that $f$ is an immersion. Recalling $0 \notin F(\mathbb R^n\setminus \{0\})$ we see that
$$f \circ p = r \circ F$$
where $r :  R^N \setminus \{0\} \to S^{N-1}, r(y) = \dfrac{y}{\lVert y \rVert}$, which is a smooth submersion. This means that the differential of $r$ has rank $N-1$ in all points. Since $F$ is an immersion, its differential has rank $n$ in all points. We conclude that the differential of $r \circ F$ has rank $\ge n-1$ in all points. This shows that the differential of $f$ must have rank $n-1$ in all points of $p(R^n \setminus \{0\}) =S^{n-1}$. Note that rank $\ge n$ is impossible for $f$.
It not yet clear whether $G(v) = G(w)$ implies $[v] = [w]$ which is needed to show that $f$ is a topological embedding.
So let $G(v) = G(w)$, i.e. $F(v) = \mu F(w)$ with $\mu = \frac{\lVert F(v) \rVert}{\lVert F(w) \rVert} > 0$. Let $\lambda = \sqrt \mu > 0$. Then
$$v_iv_j = \lambda^2 w_i w_j \text{ for all } i,j \text{ with } i \le j. \tag{2}$$
For  $x \in \mathbb R^n \setminus \{0\}$ let us consider the set $I(x) = \{ i \mid x_i \ne 0\}$. From $(2)$ we know that $v_i^2 = \lambda^2 w_i^2$ for all $i$ and therefore $I(v) = I(w) = I$. We want to show that either $v_i = \lambda w_i$ for all $i \in I$ or $v_i = -\lambda w_i$ for all $i \in I$ (which implies that $[v] = [w]$).
From $v_i^2 = \lambda^2 w_i^2$ we get $v_i = \epsilon_i \lambda w_i$ with a unique $\epsilon_i = \pm 1$ for all $i \in I$. Assume that  $\epsilon_i \ne \epsilon_j$ for some $i,j \in I$ with $i < j$. Then $v_iv_j = \epsilon_i \epsilon_j \lambda^2 w_i w_j = -\lambda^2w_i w_j$ which contradicts $(2)$. This completes our argument.
We now have proved that $f$ is a smooth embedding.
(c) Look at the last coordinate of $f(x)$. This is $G_{nn}(x)$ which is nonnegative by definition. Hence the south pole $s  =(0,\dots,0,-1)$ of $S^{N-1}$ is not contained in the image of $f$. Now use stereographic projection $\phi : S^{N-1} \setminus \{s\} \to \mathbb R^{N-1}$ to get the desired smooth embedding in form of $\phi \circ f :\mathbb RP^{n-1} \to \mathbb R^{N-1}$.
