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I have been asked to show that the function $f:\Omega \to \Omega$, $$f[x,y,z]=[x^2 +yz,y^2 ,z^2]$$ acts as a permutation on $\Omega$, the set of a one-dimensional subspaces of $V=\mathbb{F}_4^3$ (3d vector space from the field of four elements), where $[x,y,z]$ is the subspace spanned by $(x,y,z)$.

E: It is not asking to prove that $f[f[x,y,z]]=[x,y,z]$, as this is asked later.

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    Are $x,y,z$ specific elements? Or is this like an evaluation map over polynomial rings? – 2017-01-19
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    $x,y,z \in \mathbb{F_4}$ by definition, but this is just the general field with four elements. – 2017-01-19
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    Oh ok, so you've a map hopefully from $\Omega\to \Omega$, whereby you send the $1d$ space $\langle(x,y,z)\rangle\mapsto \langle(x^2+yz,y^2,z^2)\rangle$ – 2017-01-19
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    Yup that's correct – 2017-01-19
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    I've just googled "Bijective Permutation" and found "A bijective function from a set to itself is also called a permutation." from wikipedia so yeah I have to show it's bijective. Thank you very much for your help! – 2017-01-19

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For every $D\in\Omega$, let us denote by $f\left$ the set $\{f(v);v\in D\}$.

IMHO, the question asks to prove that :

  • $\forall D\in\Omega,\,f\left< D\right>\in\Omega$

  • the map $\Omega\to\Omega,D\mapsto f\left$ is bijective

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    But the function is defined to be acting on $\Omega$ i.e. $f:\Omega \to \Omega$. Surely I'd just be proving $f[x,y,z]=[x,y,z]$ – 2017-01-19