Properties of a function on $\mathbb{R}P(2)$ Define a map $F:\mathbb{R}P(2)\to \mathbb{R^3}$ such that $$F([x,y,z])=\frac{(yz,xz,xy)}{{x^2+y^2+z^2}}$$ Clearly, $F$ is well defined since $F([x,y,z])=F([\lambda x,\lambda y,\lambda z])$ where $\lambda\neq0.$ First, what I want to do is to show that $F$ is continuous. I can think of two ways. If $q:\mathbb{R^3}-\{0\}\to \mathbb{R}P(2)$ if the quotient map, and $i:\mathbb{R^3}-\{0\}\to\mathbb{R^3}$ is the inclusion, it is clear that $i=F\circ q$. Since $i$ is continuous it implies that $F$ is continuous. Alternatively, since each component of $F$ is continuous, I think it implies that $F$ is continuous. 
Now that we are done with continuity, I want to show that $F$ is injective. This follows directly from the fact that $i=F\circ q$ is injective and since $q$ is a surjection, $F$ has to be injective. 
Finally, if $||(x,y,z)||=\sqrt{x^2+y^2+z^2}$, I want to prove that $||F||$ attains a max and a min. Since $||F||:\mathbb{R}P(2)\to \mathbb{R}$ is continuous (as it is the composition of two continous maps) and since $\mathbb{R}P(2)$ is compact, $||F||$ attains a max and a min. 
 I just want to verify if these ideas are correct. I know I have left out many details ($\mathbb{R}P(2)$ is compact etc.). I will gladly provide proofs if required. I would appreciate it of anyone can spot any flaws in my arguments. Thank you!
 A: I'm going to summarize things in an answer since I have more than will fit in a comment box.
The first thing to notice is that when you argue that $F:\newcommand{\RR}{\mathbb{R}}\newcommand{\P}{\mathrm{P}}\newcommand{\RP}{\RR\P}\RP^2\to \RR^3$ is well defined, you argue (implicitly) that the function $f:\RR^3\setminus\{0\}\to\RR^3$ given by the same formula: $f(x,y,z) = \frac{(yz,xz,xy)}{x^2+y^2+z^2}$ respects the quotient relation $f(\lambda(x,y,z))=f(x,y,z)$ for $\lambda\ne 0$, hence $f$ descends to a well-defined map $F$. However, you can then notices that this also gives continuity of $F$, since $f$ is continuous (because the denominator is nonzero on $\RR^3\setminus\{0\}$). Thus continuity is almost free.
Now for injectivity, your original strategy doesn't make sense in the first place. You said $i=F\circ q$, and $i$ is injective, $q$ is surjective, so $F$ must be injective. And this would be true if the equation were true. However, $i$ being injective means that $q$ must be injective, which is certainly not the case, so this equation couldn't possibly have held. In fact, what we know is that $f=F\circ q$, and $f$ is certainly not injective because $q$ is not injective. 
In fact, $F$ is not injective. $F([0,0,1])=F([1,0,0])=(0,0,0)$.
The attaining a max and min argument however is correct. 
