I recently trying to solve the questions from John Lee's Smooth Manifold. And for the question 2.5, I gave this proof. But I felt something is not right in my proof, is there any modification that I need to fix?
Let $P:\mathbb{R}^{n+1}\setminus \{0\}\rightarrow\mathbb{R}^{k+1}\setminus \{0\}$ be a smooth function, and suppose that for some $d\in\mathbb{Z}$, $P(\lambda x)=\lambda^d P(x)$ for all $\lambda \in \mathbb{R}\setminus \{0\}$ and $x\in \mathbb{R}^{n+1}\setminus \{0\}$. Show that the map $\tilde{P}:\mathbb{RP}^n\rightarrow\mathbb{RP}^k$ defined by $\tilde{P}([x])=[P(x)]$ is well defined and smooth.
Below is my proof.
Let $x,y\in [x]$, then $y=tx$ for $t\in \mathbb{R},t\neq 0$.
Then
\begin{equation*}
\begin{aligned}
&\tilde{P}([x])=[P(x)]\\
&\tilde{P}([y])=[P(y)]=[P(tx)]=[t^dP(x)]=[P(x)]\\
\end{aligned}
\end{equation*}
Hence $\tilde{P}$ is well defined.
For the next step, note that there are natural injection from $\mathbb{R}^{n+1}$ to $\mathbb{RP}^n$ and projection from $\mathbb{R}^{k+1}$ to $\mathbb{RP}^k$, i.e,
\begin{equation*}
\begin{aligned}
i:\mathbb{RP}^n\rightarrow\mathbb{R}^{n+1}\text{ by }i([x])=x\\
\pi:\mathbb{R}^{k+1}\rightarrow \mathbb{RP}^k\text{ by }\pi(x)=[x],
\end{aligned}
\end{equation*}
And since they are injection and projection, they are both smooth functions locally around each point of x. Then by the well define of $\tilde{P}$, we obtain
\begin{equation*}
\tilde{P}([x])=(\pi\circ P\circ i)([x])
\end{equation*}
Then let $[x]\in\mathbb{RP}^n$, there is $x\in \mathbb{R}^{n+1}\setminus \{0\}$ such that $\pi_1(x)=[x]$. By smoothness of $i$, there are two smooth charts $(U',\varphi'),(U,\varphi)$ containing $x,[x]$ such that $\varphi'\circ i \circ (\varphi)^{-1}$ is smooth. Similarly, there are two charts $(V',\psi'),(V,\psi)$ containing $P(x),[P(x)]$ such that $\psi\circ\pi\circ(\psi')^{-1}$ is smooth. Hence for $x\in \varphi(U\cap \tilde{P}^{-1}(V))$, we have
\begin{equation*}
\begin{aligned}
\psi\circ \tilde{P}\circ \varphi^{-1}(x)&=\psi\circ (\pi\circ P\circ i)\circ \varphi^{-1}(x)\\
&=(\psi\circ \pi \circ (\psi')^{-1})\circ (\psi'\circ P\circ (\varphi')^{-1})\circ (\varphi' \circ i\circ \varphi^{-1})(x)\\
\end{aligned}
\end{equation*}
The first and the third part is smooth, then the second part is smooth since $P$ is smooth. Hence $\tilde{P}$ is smooth.
Any suggestion would be greatly appreciated.
