I am trying to find a map that goes from $S^2$ to $\mathbb{R}^6$ such that it is an embedding. Im trying to use the fact that $S^2$ is compact and that I can find a map $\psi$ that is an injective immersion which would then make $\psi$ an embedding. Thinking about the Monge parametrization for the sphere i would get:
$$ \psi(x,y,z) = (x,y, \sqrt{1-x^2-y^2},0,0,0) $$
whose Jacobian matrix is:
$$ J(x,y,z) = \left(\begin{array}{cccc} 1 & 0\\ 0 & 1 \\ \frac{-x}{\sqrt{1-x^2-y^2}} & \frac{-y}{\sqrt{1-x^2-y^2}} \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) $$
but this only covers the top hemisphere, I could do a composition with the negative root but I find it problematic. For instance I can't justify that the rank of the Jacobian is $3$ so I couldn't say its an immersion. Is there an injective immersion from $S^2$ to $\mathbb{R}^6$?
I don't understand the question. The standard realization of $S^2$ in $\mathbb{R}^3$ is an embedding. Adding three extra dimension does not make it any less of an embedding.