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Say I have this equation: $$f:R^3\rightarrow R^4, f(x,y,z)=(x^2-y^2,xy,xz,yz)\ $$

How does one prove from this that there exists a function $$g: RP^2 → R^4$$

I found another post that had the same question but the answer wasn't very well explained (I think anyway). So far, I have claimed that for some $a,b ,f(a) = f(b)$, given the quotient map $q: S^2 → RP^2$, $q(a) = q(b)$. $q$ is smooth and so is continuous, but I still can't understand how to actually prove $g$ exists. I mean I can't just say "here's a function g" right?

ThunderHex
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    There does exist a function $g : RP^2 \to R^4$, namely the constant function with value $0$. I suspect that this answer is not satisfactory to you, but your question puts no constraints whatsoever on the function $g$. – Lee Mosher Sep 12 '20 at 18:48
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    I think OP is looking for $f(v)=f(-v)$ for all $v\in\Bbb S^2$ and then Universal property of quoting topology https://math.stackexchange.com/questions/1493993/universal-property-in-quotient-topology – Sumanta Sep 12 '20 at 18:50
  • Sorry, I should have said f = g after q, so it is a composition, or rather, we want to find g so that this is a composition – ThunderHex Sep 12 '20 at 18:52

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The universal property of the quotient guarantees such a map. For $f$ agrees on antipodal points of the sphere.