Injective map from real projective plane to $\Bbb{R}^4$ Consider the mapping $\Bbb R^3\rightarrow\Bbb R^4$ given by
$$f(x,y,z)=(x^2-y^2,xy,xz,yz)$$
which passes to the quotient and can therefore be viewed as a map from the projective plane $\Bbb P^2=S^3/\sim$. Can we show that then the map is an injection $\Bbb P^2\hookrightarrow\Bbb R^4$ ?
 A: Each point of $\mathbb{P}^2$ has a unique representation in exactly one of the three forms
$$ (a)\ \ (1,0,0),\\
(b)\ \ (t,0,1),\\
(c)\ \ (u,1,v).$$
Under the map $f$ the points of these types map to
$$ a \to (1,0,0,0),\\
b \to (t^2,0,t,0),\\
c \to (u^2-1,u,uv,v).$$
To check the overall map is injective, we need to check whether (i) two different points from the same form can map to the same point under $f$, and whether (ii) one point from one of the forms can map to the same as one point of another form.
Check of (i)
There is only one point of type (a), nothing to check here. Two points of type (b) have the same $t$ value from the third coordinate of the b image under $f$, and two points of type (c) have both the same $u$ and the same $v$ if their images map, from the second and fourth coordinates of the image of c types.
Check of (ii)
If the image of the (only) type (a) matches an image of a type (b) then $t=0$ from the third coordinates, but then the (b) point image has first coordinate zero and doesn't match the (a) point image first coordinate $1.$ 
Next, if the image of type (a) were to match an image of a type (c), then $u=v=0$ on comparing second and fourth coordinates of the images, but then the first coordinate of the (c) image would be $-1,$ not matching the first coordinate of the (a) image.
Lastly, if the image of a type (b) were to match the image of a type (c) then $u=v=0$ would follow on comparison of second and fourth coordinates of the images, so that since that would make the third coordinate of the (c) image zero we would have also $t=0$ to make third coordinates match, but then the first coordinates of the (b) and (c) images would be $0$ and $-1$ respectively.
This finished the (tedious) checking, and we can conclude the map is indeed injective into $\mathbb{R}^4$ from the domain $\mathbb{P}^2.$
Added Later:
The following more "bare bones" approach is added, because in the previous treatment only particular representations of points of $\mathbb{P}^2$ werre shown to map under $f$ to distinct points of $\mathbb{R}^4.$ That may actually be enough, however I couldn't see why, hence this "add on".
Let the coordinates of $f(x,y,z)$ be 
$$x^2-y^2=a,\\ xy=b,\\ xz=c,\\ yz=d.$$
We want to show these coordinates determine $(x,y,z)$ as a point of $\mathbb{P}^2.$ This means we need to determine which if any of $x,y,z$ are zero, and in case there are more than one nonzero among $x,y,z$ to determine the value(s) of the ratios from $a,b,c,d.$
Now we do not have all of $x,y,z$ zero in $\mathbb{P}^2.$ We treat first when two of $x,y,z$ are zero, the other nonzero. [These represent single points in $\mathbb{P}^2.$] This case occurs iff $b=c=d=0,$ and which case it is can be decided by the sign of $a$. The case $(0,0,z)$ has $a=0.$ The case $(0,y,0)$ has $a=-y^2<0,$ while the case $(x,0,0)$ has $a=x^2>0.$
Next we look at when one of $x,y,z$ is zero ande the other two nonzero. This occurs iff two of $b,c,d$ are zero and the third nonzero, and we can tell the $x,y,z$ situation from these. 
We have $x=0$ and $y,z$ nonzero iff $b=c=0, d \neq 0.$ in this case from $yz=d$ and $-y^2=a$ we get $y/z=-a/d$, so that the point $(0,y,z)$ is determined in $\mathbb{P}^2.$
We have $y=0$ and $x,z$ nonzero iff $b=d=0, c\neq 0.$ Then from $xz=c$ and $x^2=a$ we have $x/z=a/c,$ determining the point $(x,0,z)$ in \mathbb{P}^2.$
The third case of two nonzero is the most involved: we have $z=0$ and $x,y$ nonzero iff $c=d=0,b\neq 0.$ The equations now are $x^2-y^2=a$ and $xy=b$ where $b \neq 0$. Writing $y=b/x$ and putting $t=x^2$ we get to $t-b^2/t=a$ which has only one positive solution for $t=x^2,$ namely
$$t=\frac{a+\sqrt{a^2+4b^2}}{2}.$$
Then from $y=b/x$ we have $y/x=b/x^2=b/t$  with $t$ the above expression in $a,b,$ so that here again the point $(x,y,0)$ is determined in $\mathbb{P}^2.$
The last (and easiest) case is when $x,y,z$ are each nonzero, which occurs iff $b,c,d$ are each nonzero. In this case, $x/z=b/d$ and $y/z=b/c,$ determining the point $(x,y,z)$ in $\mathbb{P}^2.$
