Let be $X=\{(x,y,z,t) | x^2+6y^2+4z^2+t^2=1 \}$. I have proved that $X$ is a submanifold of $\mathbb{R}^4$ of dimension $3$. I have to prove that $X$ is compact and connected.
My idea, thinking of the ellipsoid in dimension $2$, is to prove that $X$ is homeomorphic to the sphere $S^3$.
The candidate homeomorphism that I tried to write down is: $$\varphi: S^3 \rightarrow X$$ $$ (x,y,z,t) \rightarrow \frac{1}{\sqrt{x^2+6y^2+4z^2+t^2}} (x,y,z,t)=\frac{1}{\sqrt{1+5y^2+3z^2}} (x,y,z,t).$$
(obtained from associating a point $\bar x$ on $R^3$ to the point given by the intersection between $X$ and line throuhg $\bar x $ and the origin).
The problem is that the inverse is : $$ (x,y,z,t) \rightarrow \frac{1}{\sqrt{1-3z^2-5^2}} (x,y,z,t),$$ that have problems with the denominator.
Is this way of proceeding incorrect? How else can I do?
Thanks for the help!
Hint For $a, b, c, d$ all nonzero, the diagonal maps $\Bbb R^4 \to \Bbb R^4$ defined by $$(x, y, z, t) \mapsto (ax, by, cz, dt)$$ are invertible linear transformations and hence continuous with continuous inverse.