Finding a homeomorphism between $FP^n$ and $S^n_F/{\sim}$ I need help in this:
Let $F$ is $\mathbb{R},\mathbb{C},H$ ($H$ means the quaternions).
Prove that the projective space $PF^n$
$$ PF^n= \{ F^{n+1}\backslash \{0\})/\sim 
|  x{\sim}\lambda x , \lambda \neq 0\in{F} \}$$ is homeomorphic to the quoitent space $SF^{n}/{\sim}$ under the proportion equivalence $x\sim\lambda x$ for $\lambda\in{S^1_F}$.
Where
$S^n_{F}={x\in{F^{n+1}}: ||x||=1}$
 A: In this forum you will find several similar questions, but usually only for $F  = \mathbb R$.
So let $p : F^{n+1} \setminus \{0\} \to PF^n$ denote the quotient map. Define $r : F^{n+1} \setminus \{0\} \to S^n_F, r(x) = \frac{x}{\lVert x \rVert}$. This map is a  retraction.
Consider the restriction $q : S^n_F \to PF^n$.

*

*We have $q \circ r = p$. In fact, for each $x \in F^{n+1}$ we have $x \sim r(x) \in S^n_F$. Thus $q(r(x)) = p(r(x)) = p(x)$.


*$q$ is surjective. This follows from 1. since each $\xi \in PF^n$ has the form $\xi = p(x)$ which implies $\xi = q(r(x))$.


*We have $q(x) = q(y)$ if and only if $x = \lambda y$ for some $\lambda \in S^1_F$. The "if" part is trivial. So let $q(x) = q(y)$ which is the same as $p(x) = p(y)$, Hence there exists $\lambda \in F \setminus \{0\}$ such that $x =  \lambda y$. This implies $1 = \lVert x \rVert = \lVert \lambda y \rVert = \lvert \lambda \rvert \lVert x \rVert =  \lvert \lambda \rvert$, i.e. $\lambda \in S^1_F$.
It remains to show that $q$ is a quotient map, i.e. that $U \subset PF^n$ is open if and only if $q^{-1}(U) \subset S^n_F$ is open. Since $q$ is continuous, the "only if" part is trivial. So let $q^{-1}(U) \subset S^n_F$ be open. Hence $r^{-1}(q^{-1}(U)) \subset F^{n+1} \setminus \{0\}$ is open. But $r^{-1}(q^{-1}(U)) = (q \circ r)^{-1}(U) = p^{-1}(U)$, thus $p^{-1}(U)$ is open in $F^{n+1} \setminus \{0\}$. This implies that $U$ is open in $PF^n$.
