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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~

WishingFish
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    You've shown that the locus $p^{-1}(X_d)\subset\mathbb R^3-{0}$ is a smooth surface. You haven't yet shown $X_d$ is a smooth curve in $\mathbb R^2$. You have a gap: Yes, the projection to $T_y\mathbb RP^3$ is surjective, but why is its kernel not contained in $T_x\big(p^{-1}(X_d)\big)$? (For future reference, check out Euler's theorem on homogeneous functions.) – Ted Shifrin Jul 04 '13 at 12:48
  • Thanks a lot @TedShifrin - but why do I need to show that $X_d$ is a smooth curve in $\mathbb{R}^2$? I am only asked to show $X_d$ is a submanifold? – WishingFish Jul 04 '13 at 17:56
  • AGH. That was a typo. You need to show $X_d\subset\mathbb RP^2$ is a smooth curve. So you either need to work in local coordinates on $\mathbb RP^2$ [after all, $f$ is not a function on $\mathbb RP^2$] or you need to fix your regular value argument in $\mathbb R^3-{0}$. In fact, $\ker dp_x$ is contained in $T_x\big(p^{-1}(X_d)\big)$ (because $p$ collapses the lines through the origin to points). So you still have a problem to deal with: If $Y\subset M$ is a manifold and $p\colon M\to N$ is a submersion, do we know that $p(Y)$ is always a submanifold of $N$? – Ted Shifrin Jul 04 '13 at 18:18

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