One-to-one mapping from $\mathbb{R}^4$ to $\mathbb{R}^3$ I'm trying to define a mapping from $\mathbb{R}^4$ into $\mathbb{R}^3$ that takes the flat torus to a torus of revolution. 

Where the flat torus is defined by $x(u,v) = (\cos u, \sin u, \cos v, \sin v)$.
And the torus of revolution by $x(u,v) = ( (R + r \cos u)\cos v, (R + r \cos u)\sin v, r \sin u)$.

I think an appropriate map would be:
$f(x,y,z,w) = ((R + r x)z, (R + r x)w, r y)$ where $R$, $r$ are constants greater than $0$.

But now I'm having trouble showing this is one-to-one.
Compute the Jacobians of the two mappings.  I'll call your first mapping $X(u, v)$ so as not to re-use the variable $x$.  Let $R &gt; r &gt; 0$.

$$\frac{\partial X}{\partial(u, v)} = \left[ \begin{array}{cc}
-\sin u &amp; 0 \\ \cos u &amp; 0 \\ 0 &amp; -\sin v \\ 0 &amp; \cos v \end{array} \right] $$

$$\frac{\partial f}{\partial(x, y, z, w)} = \left[ \begin{array}{cccc}
rz &amp; 0 &amp; R + rx &amp; 0 \\ rw &amp; 0 &amp; 0 &amp; R + rx \\ 0 &amp; r &amp; 0 &amp; 0 \end{array} \right] $$

The embedding in $\mathbb{R}^3$ is the composite $f \circ X$, whose Jacobian is the product of the Jacobians:

$$ \frac{\partial f}{\partial(x,y,z,w)}\frac{\partial X}{\partial(u,v)} 
= \left[ \begin{array}{cc}
-rz\sin u &amp; -(R + rx)\sin v \\ -rw\sin u &amp; (R+rx)\cos v \\ r\cos u &amp; 0  \end{array} \right]
= \left[ \begin{array}{cc}
-r\cos v\sin u &amp; -(R + r\cos u)\sin v \\ -r\sin v\sin u &amp; (R+r\cos u)\cos v \\ r\cos u &amp; 0  \end{array} \right]

$$

(I used sage notebook www.sagenb.org for my computations.)  

Now all of the questions concerning 1-to-1 can be phrased in terms of rank and nullity of these matrices.  For example, it easy to see the first and third matrices have rank 2 (so nullity=0) for all choices of $u, v$, (which shows both representations of the torus are immersive, meaning an appropriate restriction of domain will make those maps 1-to-1).

Now $f$ is 1-to-1 when restricted to $\mathrm{im} X$ (the flat torus) because $f \circ X$ is immersive and by restricting to $u, v \in [0, 2\pi)$, both $f$ and $f \circ X$ become 1-to-1.  In other words, if $f( X(u,v) ) = f(X(u&#39;, v&#39;))$, then $X(u,v) = X(u&#39;, v&#39;)$.

Hope this helps!
Do you need to explicitly give the map or are you asked to prove both sets have the same cardinilty? 

If not, it is a good exercise to prove $\mathbb{R}$ and $\mathbb{R}^2$ are both the same size (Schroeder-Bernstein theorem - consider decimal expansions to show a surjection) then via induction and the fact both sets are subsets of $\mathbb{R}^n$ we see there is a bijection between them. Explicitly finding a bijection is not clear to me, perhaps it is to someone else.