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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.

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Presumably the mapping is only going to be one-to-one when restricted to the torus? – JSchlather Nov 30 '11 at 2:55

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 $R > r > 0$.

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

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

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

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

(I used sage notebook 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 $f( X(u,v) ) = f(X(u', v'))$, then $X(u,v) = X(u', v')$.

Hope this helps!

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Shaun's answer is insufficient since there are immersions which are not 1-1. For example, the figure 8 is an immersed circle. Also, the torus covers itself and all covering maps are immersions.

Your parametrization of the torus of rotation is the the same as in You just have to notice that the minimal period in both coordinates of the $uv$ plane is the same $2\pi$ in the case of both the flat and rotated tori.

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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.

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It doesn't look like you read the question very carefully... – JSchlather Nov 30 '11 at 2:54
It's possible that he only read the title of the question. – Paul Nov 30 '11 at 2:57

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