The Charts on the Projective Space of a Real Vector Space Let $V$ be an $n+1$-dimensional normed real vector space. We define $P(V)$ as the topological space $(V-\{0\})/\sim$ where $\sim$ is an equivalence relation on $V-\{0\}$ which identifies $v$ and $tv$ for each $v\in V-\{0\}$ and $t\in \mathbf R-\{0\}$.

I want to show that $P(V)$ is a topological $n$-manifold.

One way I tried how to do it is by taking an $n$-dimensional subspace $U_n$ and a $1$ dimensional subspace $U_1$ of $V$ as define $\varphi:V-U_n\to U_n$ as $\varphi(v)=p(v/\|v\|)$, where $p$ is the projection on $U_n$ with respect to $U_1$. But this doesn't even respect the equivalence relation $\sim$.
Of course, another way to do this would to take a linear isomorphism of $V$ with $\mathbf R^{n+1}$ then then use the charts on $\mathbf R\mathbf P^n$ to construct charts on $P(V)$.
I want to do it in an intrinsic manner---without making any reference to $\mathbf R^{n+1}$, with my main motivation being to understand tha Grassmannians better.
Can somebody help me with these charts?
Thanks.
 A: The standard charts are $$U_i = \{ [x_1, \dots x_{n+2}] \in \Bbb P (V) \space | \space x_i \neq 0 \} ,$$ with $$h_i : U_i \to \Bbb R ^{n+1}, \space h_i([x_1, \dots, x_{n+2}]) = (\frac {x_1} {x_i}, \dots , \frac {x_{i-1}} {x_i}, \frac {x_{i+1}} {x_i}, \dots, \frac {x_{n+2}} {x_i})$$ for $i=1, \dots, n+2$ (note the "skipped" coordinate). Necessarily they cover $\Bbb P (V)$ since there is no point in it having all projective coordinates $0$. Each of the open charts is thus diffeomorphic (homeomorphic, etc.) to $\Bbb R ^{n+1}$.
Later edit:
In order not to explicitly use coordinates, one might mimic the above construction in the following way. Pick $\omega \in V^* \setminus \{0\}$ and let $H _\omega = \ker \space \omega$. Let $\pi : V \to \Bbb P (V), \space \pi(v) = [v]$ be the natural projection. Let $U_\omega = \pi (V \setminus H_\omega)$ and define $h _\omega : U _\omega \to H _\omega, \space h _\omega([v]) = {\rm proj} _{H _\omega} \frac v {\omega(v)}$ (the projection of $\frac v {\omega(v)}$ onto $H _\omega$; this projection is defined by considering the direct sum decomposition $V \simeq H_\omega \oplus {\rm range} \space \omega$). Let us show that $(U_\omega, h_\omega)$ is a chart.
First, note that $\pi ^{-1} (U_\omega) = V \setminus H_\omega$. If this were not true, then you could find $v \in V \setminus H_\omega$ and $h \in H _\omega$ with $\pi (h) = [v]$, but this would mean that $v$ is a non-zero multiple of $h$ and thus belongs to $H_\omega$, which is a contradiction. Therefore, $\pi ^{-1} (U_\omega) = V \setminus H_\omega$ and since $V \setminus H_\omega$ is open, the way topology on $\Bbb P (V)$ is defined (it is the final topology) implies that $U_\omega$ is open too.
Next, note that the fraction $\frac v {\omega(v)}$ onto $H _\omega$ is homogeneous of degree $0$, therefore it only depends on $[v]$, not on $v$, so $h_\omega$ is well-defined. It is continuous, too, as a composition of continuous maps. Bijectivity follows easily. That the inverse is also continuous seems obvious to me, but I find it annoying to write down all the details.
Since $\Bbb P (V) = \bigcup \limits _{\omega \in V^* \setminus \{0\}} U_\omega$, each point of $\Bbb P (V)$ has a local chart of this type around it (and all of them form an atlas, but this too needs checking, that I shall not do).
All this does not prove yet that $\Bbb P (V)$ is a topological manifold, because each $h_\omega$ takes values an a different $H_\omega$. You should finally show that $H_\omega \simeq \Bbb R ^n$ (topologically), but guess what? In order to show that, you will now need to resort to coordinates - but at least this only happens at the very end of the proof! In any case, no proof of what you ask can be completely coordinates-free.
A: For each basis $\Gamma = (\lambda_1, \cdots, \lambda_{n+1})$ of the dual space $V^*$ let $W \subset P(V)$ defined by $W := \{ [w] : \lambda_{n+1}(v) \neq 0 \}$. Then $W$ is an open subset of $P(V)$. Declare $x_1 := \frac{\lambda_1}{\lambda_{n+1}}, \cdots, x_n := \frac{\lambda_n}{\lambda_{n+1}}$ coordinates  $(x_1,\cdots,x_n)$ of $W$. Then it is not difficult to check that coordinates changes are smooth hence $P(V)$ is a topological manifold. \
If you do not want to use basis of the dual space here is a possible solution.
You have to assume that the unit sphere $S:= \{v : \| v \| = 1 \}$ of your normed space is a topological manifold. Then the natural projection $\pi: S \to P(V) $ is a so called double covering hence you can (locally) pull-back charts of $S$ and so obtain that $P(V)$ is a topological manifold.
