Smoothness of a map between $\Bbb {RP}^n$ and $\Bbb{RP}^k$ I would like to prove the following statement.
If $P: \mathbb{R}^{n+1} \backslash \{0\} \to \mathbb{R}^{k+1} \backslash \{0\}$ is smooth and homogeneous of degree $d$, then the map $\widetilde P: \mathbb{RP}^n \to \mathbb{RP}^k$ defined by $\widetilde P([x]) = [P(x)]$ is smooth.
If two charts $\varphi, \psi$ are given in $\mathbb{RP}^n$ and $\mathbb{RP}^k$, respectively, I get the expression $\varphi \circ \widetilde P \circ \psi^{-1} (x_1, ..., x_n) = \varphi \circ [P(x_1, ... ,x_{i-1},1,x_{i_1}, ... x_n)]$, but am unable to make use of the smoothness of $P$ at this point. Any ideas about this?
 A: Let $P(x_0, \dots, x_n) = (P_0(x_0, \dots, x_n), \dots, P_k(x_0, \dots, x_n))$ where $P_0, \dots, P_k : \mathbb{R}^{n+1} \to \mathbb{R}$ are smooth homogeneous polynomials of degree $d$. 
If you had done this, I think, based on what you had already done, you would have been able to complete the proof. Just in case, I have included the proof below.

Let $U_i = \{[x_0, \dots, x_n] \in \mathbb{RP}^n \mid x_i \neq 0\}$ and $\varphi_i : U_i \to \mathbb{R}^n$, given by 
$$\varphi_i([x_0, \dots, x_n]) = \left(\frac{x_0}{x_i}, \dots, \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \dots, \frac{x_n}{x_i}\right)$$
with inverse $\varphi_i^{-1} : \mathbb{R}^n \to U_i$ given by $\varphi_i^{-1}(x_1, \dots, x_n) = (x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)$.
Let $V_j = \{[y_0, \dots, y_k] \in \mathbb{RP}^k \mid y_j \neq 0\}$ and $\psi_j : V_j \to \mathbb{R}^k$, given by
$$\psi_j([y_0, \dots, y_k]) = \left(\frac{y_0}{y_j}, \dots, \frac{y_{j-1}}{y_j}, \frac{y_{j+1}}{y_j}, \dots, \frac{y_k}{y_j}\right)$$
with inverse $\psi_j^{-1} : \mathbb{R}^k \to V_j$ given by $\psi_i^{-1}(y_1, \dots, y_k) = (y_1, \dots, y_{j-1}, 1, y_{j+1}, \dots, y_k)$.
For each $i$ and $j$ such that $U_i\cap\widetilde{P}^{-1}(V_j) \neq \emptyset$, we have the map 
$$\psi_j\circ\widetilde{P}\circ\varphi_i^{-1} : \varphi_i^{-1}(U_i\cap\widetilde{P}^{-1}(V_j)) \to \psi_j(\widetilde{P}(U_i)\cap V_j)$$
given by
\begin{align*}
&(\psi_j\circ\widetilde{P}\circ\varphi_i^{-1})(x_1, \dots, x_n)\\ 
=&\ \psi_j(\widetilde{P}(\varphi_i^{-1}(x_1, \dots, x_n)))\\
=&\ \psi_j(\widetilde{P}(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n))\\
=&\ \psi_j(P_0(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n), \dots, P_k(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n))\\
=&\ \left(\frac{P_0(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}{P_j(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}, \dots, \frac{P_{j-1}(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}{P_j(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}, \frac{P_{j+1}(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}{P_j(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}, \dots, \frac{P_k(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}{P_j(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n)}\right).
\end{align*}
As $(x_1, \dots, x_n) \in \varphi_i^{-1}(U_i\cap\widetilde{P}^{-1}(V_j))$, $P_j(x_1, \dots, x_{i-1}, 1, x_i, \dots, x_n) \neq 0$, so $\psi_j\circ\widetilde{P}\circ\varphi_i^{-1}$ is smooth (rational functions are smooth on the complement of the zero locus of the denominator).
Therefore, $\widetilde{P} : \mathbb{RP}^n \to \mathbb{RP}^k$ is smooth.

An almost identical argument can be used to show an analogous result in the complex setting:

If $P : \mathbb{C}^{n+1} \to \mathbb{C}^{k+1}$ is a degree $d$ homogeneous polynomial mapping, then $\widetilde{P} : \mathbb{CP}^n \to \mathbb{CP}^k$ given by $\widetilde{P}([z]) = [P(z)]$ is holomorphic.

