Prove that $g$ is a submanifold: $g (t,u,v) = (t^2,u^2,v^2,\sqrt{2}uv, \sqrt{2}tv,\sqrt{2}tu)$ We consider $g : (t,u,v)\in \mathbb{R}^3 \mapsto (t^2,u^2,v^2,\sqrt{2}uv, \sqrt{2}tv,\sqrt{2}tu)\in\mathbb{R}^6$. I have to prove that $g(\mathbb{S}^2)$ is a submanifold of $\mathbb{R}^6$.
$dg_{(t,u,v)}=\begin{pmatrix}2t&0&0\\ 
0&2u&0\\ 
0&0&2v\\ 
0&\sqrt{2}v&\sqrt{2}u\\
\sqrt{2}v&0&\sqrt{2}t\\
\sqrt{2}u&\sqrt{2}t&0\\
\end{pmatrix} $
So it proves that $g$ is an immersion (the rank is $3$ clearly) for $(\mathbb{R}^3-\{(0,0,0)\})$.
Now how to prove that $\mathbb{S}^2$ is homeomorphic to $g(\mathbb{S}^2)$ ? I must prove that $g$ is injective  ?
Thanks in advance !

In  fact, the map $g$ is $\bf not$ injective, as $g(u,v,w)=g(-u,-v,-w)$. It defines a map from the real projective space $\bf RP^2$ to $\bf R^6$.
$\bf RP^2$ is defined as the quotient of the sphere by the involution  $I(u,v,w)=-(u,v,w)$ 
It is easy to see that $g(u,v,w)=g(u',v',w')$ iff $(u,v,w)=(u',v',w')$ or $I(u,v,w)=(u',v',w')$, therefore the map $g$ descend to an injective map $\bar g$ from $\bf RP^2$, which is an homeomoprhims to its image.
Then one can say that there exixts a unique structure of a surface on $\bf RP^2$ such that the natural projection is differentiable and a local diffeomorphism, so in fact $\bar g$ is an immersion and an homemorphism onto its image therefore a diffeomorphism
