I'm working on a question that consists of two parts.
Let $\pi : \mathbb{R}^{n+1} \backslash \{0\} \to \mathbb{R}\mathbb{P}^n$ be the natural projection. Let $P(x_0,\ldots,x_n)$ be a homogenenous polynomial such that $$ \left( \frac{\partial P}{x^0}, \ldots, \frac{\partial P}{\partial x^n} \right)\mid_p \neq (0,\ldots,0). $$ outside $0 \in \mathbb{R}^{n+1}$. Define a structure of a smooth manifold on the set $\pi(\{P = 0\})$.
Prove that in the case $n = 2$ and $\deg P = 3$, the resulting manifold is either homeomorphic to $S^1$ or $S^1 \sqcup S^1$, and that both options can be realized.
I have attempted part (1), is this correct? But for part (2) I have no idea how to prove that, can someone outline a proof strategy here? I'm new to differential geometry so I have only had elementary definitions and examples so far. This is my attempt for (1):
Define $M = \pi(P^{-1}(0)) \subset \mathbb{R}\mathbb{P}^n$. Firstly, we show that this is a well-defined subset of the projective plane. Indeed, for any $[x^0 : \ldots : x^n] \in \mathbb{R}\mathbb{P}^n$, if we have $P(x^0,\ldots,x^n) = 0$, then since $P$ is homogeneous (say of degree $m$), we have for any scalar $\lambda \neq 0$ that $P(\lambda x^0,\ldots, \lambda x^n) = \lambda^m P(x^0,\ldots,x^n) = 0$. So the property is well-defined on equivalence classes.
We claim that $M$ is an $(n-1)$-dimensional manifold. We construct the charts as follows. Recall that
$$
U_j = \{[x^0 : \ldots : x^n] ; x^j \neq 0\}.
$$
Now for every $(x^0,\ldots,x^n) \in \pi^{-1}(U_j)$, we consider the point
$$
\vec{p} = \left( \frac{x^0}{x^j},\ldots,\frac{x^{j-1}}{x^j},1,\frac{x^{j+1}}{x^j},\ldots,\frac{x^n}{x_j} \right) \in \mathbb{R}^{n+1}
$$
By the assumption, there is some $i$ for which $\frac{\partial P}{\partial x^i}(\vec{p}) \neq 0$. Hence, by the same trick as in the previous question, using the implicit function theorem there exist open sets $U_{i,j} \subset \mathbb{R}^n$ and $V_{i,j} \subset \mathbb{R}$ with $\vec{p} \in U_{i,j} \times V_{i,j}$ (neighborhood of $\vec{p}$) and smooth map $\Phi_{i,j} : U_{i,j} \to V_{i,j}$ such that for every $\vec{x} \in P^{-1}(0) \cap (U_{i,j} \times V_{i,j})$ we have that $x^i = \Phi(x^1,\ldots,\widehat{x^i},\ldots,x^n)$. In particular for $\vec{p}$ we have
$$
p^i = \frac{x^i}{x^j} = \Phi(\frac{x^0}{x^j},\ldots,\frac{\widehat{x^i}}{x^j},\ldots,\frac{x^n}{x^j}).
$$
Since $\pi$ is an open map, we obtain an open set $W_{i,j} := \pi(U_{i,j} \times V_{i,j})$ in $\mathbb{R}\mathbb{P}^n$, and we have that $[\vec{x}] = x^j [\vec{p}] \in W_{i,j}$. We define the coordinate map, assuming $i < j$,
$$
\varphi : W_{i,j} \to \mathbb{R}^{n-1} : \vec{x} \mapsto \left( \frac{x^0}{x_j},\ldots,\widehat{\frac{x^i}{x^j}},\ldots,\widehat{\frac{x^j}{x^j}},\ldots,x^n \right).
$$
with inverse
\begin{align*}
\varphi^{-1} : \mathbb{R}^{n-1} \to W_{i,j} :
& (u^0,\ldots,u^{n-2}) \mapsto \\
& [u^0 : \ldots : u^{i-1} : \Phi_{i,j}(u^0,\ldots,u^{j-1},1,u^j,\ldots,u^{n-2}) : u^i : \ldots : u^{j-1} : 1 : u^j : \ldots : u^{n-2}].
\end{align*}
Question, what if $i = j$ in this construction? It does not seem to work in this case.