I am trying to prove that the map $$F:(x,y,z)\in S^2\mapsto (x^2-y^2,xy,xz,yz)\in\mathbb{R}^4$$ induces a $C^\infty$ map $G:\mathbb{RP}^2\rightarrow\mathbb{R}^4$ which is an embedding.
What I would do is:
- Define $G: [x,y,z]\in\mathbb{RP}^2\mapsto F(x,y,x)\in\mathbb{R}^4$. This is well defined beacuse $F$ is both adequate to the antipodal quotient relation and is homogeneous of deegre two component-wise.
- To show that it is $C^\infty$ I would observe that the projection on the quotient $\pi$ is a local diffeomorphism for the usual structures on the sphere and on $\mathbb{RP}^2$, so is $F$, being the restriction of a $C^\infty(\mathbb{R}^3,\mathbb{R}^4)$ map to a subvariety, and hence so is $G$ since it holds $G\circ\pi=F$.
- To show $G$ is an immersion I would show that the differential $F_*:T_p\mathbb{R}^3\rightarrow T_p\mathbb{R^4}$ has full rank on the points of $S^2$: then being $S^2$ a subvariety its restriction to $T_pS^2$ this will coincide with the differential of $F|_{S^2}$, which is what we search for. Then one observes $G_*=F_*(\pi^{-1})^*$ and again uses the fact $\pi$ is a local diffeomorphism. This is the point I am more uncertain of.
- Finally to show $G$ is an homeomorphism onto its image one exploits usual results about quotient topology plus the fact that $S^1$ and hence $\mathbb{RP}^2$ is compact and $G(\mathbb{RP}^2)\subset\mathbb{R}^4$ is Hausdorff.
Is the outline of the solution correct? Thanks in advance
About point 1. : proving that it is antipode-invariant is actually enough, since $S^2$ mod the antipodal map is $\mathbb{RP}^2$
About point 2. : using that it's a local diffeomorphism is essentially reproving the defining property of the quotient; so you don't actually need to do it (although of course, you can!)
About point 3. : This is correct; the fact that it's a local diffeomorphism implies that $d_p\pi$ is an isomorphism on tangent spaces, so you get full rank for $d_pG$ as well.
About point 4. : everything is correct, except that you probably meant $S^2$ and not $S^1$.