Show that the function $f: RP^2 \to R^4$ given by $\{x,-x\} \mapsto (x_1^2-x_2^2,x_1x_2,x_1x_3,x_2x_3)$ is injective. I am writing up a solution to the exercises of Czes Kosniowski's A First Course In Algebraic Topology, and I am stuck on this problem among others. I have tried a direct approach to the proof, using the traditional definition of injectivity as well as the contrapositive version, but I keep running into dead ends. A point in the right direction would help.
 A: First off, I think that probably the right notation would be $\{(x_1, x_2, x_3),(-x_1,-x_2,-x_3)\} \mapsto \dots.$ My solution has a "corner case" that I'll point out and leave to you to prove. So suppose that two point $x=\{(x_1, x_2, x_3),(-x_1,-x_2,-x_3)\}, y=\{(y_1, x_2, x_3),(-y_1,-y_2,-y_3)\}$ map to the same point, so that $x_1^2-x_2^2=y_1^2-y_2^2$, etc. 
Since the points are on a sphere, $x_1^2+x_2^2+x_3^2 (=1)=y_1^2+y_2^2+y_3^2$. Add to this equation $x_1^2-x_2^2=y_1^2-y_2^2$ and complete the square to get $(\sqrt 2 x_1+x_3)^2-2\sqrt 2 x_1x_3 = (\sqrt 2 y_1+y_3)^2-2\sqrt 2 y_1y_3$. Since $x_1x_3=y_1y_3$, the squares are equal, so $(\sqrt 2 x_1+x_3) = \pm \sqrt 2 y_1+y_3)$. Notice that the left side of this equation times $x_2$ is equal to plus or minus the right side times $y_2$, so as long as $\sqrt 2 x_1+x_3\neq 0$ we conclude that $x_2=\pm y_2$. Then $x_1x_2=y_1y_2, x_2x_3=y_2y_3$ give us the other two appropriate equalities.
If $\sqrt 2 x_1+x_3=0$ start over, but subtract $x_1^2-x_2^2=y_1^2-y_2^2$ instead of adding it, and go through the same steps, provided $\sqrt 2 x_2+x_3\neq 0$.
The corner case is where both $\sqrt 2 x_1+x_3$ and $\sqrt 2 x_2+x_3$ are $0$, but then $x_1=x_2$ and it's pretty easy to finish!
