I am wondering if my proof is legit? The ending looks rather soft. I don't know whether it is correct, or how to rephrase it if it is correct.
Let $p: \mathbb{R}^3 - \{0\} \to \mathbb{R}P^2$ be the usual projection. Prove that the solution set in $\mathbb{R}^3 - \{0\}$ of $x^d +y^d= z^d$ has the form $p^{-1}(X_d)$ for some subset $X_d$ of $\mathbb{R}P^2$.
In plain English, I guess I am asked to show that the solution set in $\mathbb{R}^3 - \{0\}$ is the preimage of some subset of $\mathbb{R}P^2$. Since the subset of $\mathbb{R}P^2$ a identified antipodal pair of points contracted from the line passing these points, $p^{-1}(X_d)$ is just a set of lines passing through origin.
But clearly, consider a point $(a,b,c)$ in the solution set, clearly, $\lambda(a,b,c)$ is also in the solution set $\forall \lambda \in \mathbb{R} - \{0\}.$ This collides with what described above.
Prove that $X_d$ is a manifold. (Hint: Local coordinates.)
The hint looks very disturbing to me because I didn't use it at all. I sense there may be something wrong but I couldn't see it.
I want to show that 0 is the regular value of $x^d +y^d - z^d =0$. In other words, $\forall x: f(x)=0, df_x = d x^{d-1} + dy^{d-1} - dz^{d-1}$ must be surjective. Since it is given that $x,y,z$ can't equal to 0 together, $df_x: T_x(\mathbb{R}^3 - \{0\}) \to T_y(\mathbb{R}P^2)$ is surjective. Hence by preimage theorem, the preimage of 0, which is the solution set $X_d$ is a submanifold.
Thank you~