Smoothness of $f : S^1 \times S^1 \to T$ Given $f : S^1 \times S^1 \to T \subset \mathbb{R}^3$ such that $$(u,v)  \mapsto (u_1 (R+ r \ v_1), \ u_2 (R+r \ v_1), \ r \ v_2)$$
where $S^1$ is the unit-circle and $T$ is the familiar ring torus with $R > r > 0$
The book I am studying from mentions that $f$ is a diffeomorphism, but I don't understand its smoothness at all.
I know that both $S^1 \times S^1$ and $T$ are two-dimensional and if $f$ is a diffeomorphism, then $Df$ is invertiable, so it must be a $2\times2$ matrix.  But the map seperates the components of $u$ and $v$, so I am not sure how the Jacobian looks like. If I consider each components $u_i$ and $v_i$ seperately, then it's $3\times4$ and is not invertiable.
Any clarification would be great!
 A: Although the domain $S^1 \times S^1$ is a submanifold of 4-dimensional space $\mathbb R^4$, and the range $T$ is a submanifold of the 3-dimensional space $\mathbb R^3$, it does not follow that $Df$ is a $3 \times 4$ matrix. 
Given a point $p \in S^1 \times S^1$ and its image point $q = f(p) \in T$, the domain of the linear transformation $D_p f$ is a 2-dimensional vector space, namely the 2-dimensional subspace $T_p(S^1 \times S^1) \subset T_p \mathbb R^4 \approx \mathbb R^4$. Also, the range is a 2-dimensional vector space, namely the 2-dimensional subspace $T_q(T) \subset T_q \mathbb R^3 \approx \mathbb R^3$. The statement that $f$ is a diffeomorphism implies that the linear transformation 
$$D_p f : T_p(S^1 \times S^1) \to T_q(T)
$$ 
is an isomorphism between those two 2-dimensional vector spaces. So far there is no matrix involved, because the domain and range vector spaces of $D_p f$ have not been assigned any particular basis.
If you want to express $D_p f$ as a $2 \times 2$ matrix, then you have to choose coordinate charts for $S^1 \times S^1$ around $p$ and for $T$ around $q$, which will in turn determine bases for the domain and range vector spaces. You could, for instance, choose $2$ out of the $4$ coordinates of $\mathbb R^4$ as coordinates for $S^1 \times S^1$ in a neighborhood of $p$, but the choice of those two coordinates must be made carefully. Not just any 2 out of the 4 will work: you must choose those $2$ so that the projection of the 2-dimensional subspace $T_p(S^1 \times S^1)$ onto the coordinate subspace of those 2 coordinates is an isomorphism. Similarly, you can choose $2$ out of the $3$ coordinates of $\mathbb R^3$ as coordinates for $T$ in a neighborhood of $q$, but again not just any 2 out of the 3 will work. 
Once you have chosen the correct coordinates in both the domain and range in this fashion, then yes, you'll get a $2 \times 2$ invertible matrix.
A: To see that $f$ is smooth is very easy. Define $F: \mathbb R^4 \to \mathbb R^3, F(u_1,u_2,v_1,v_2) = (u_1 (R+ r \ v_1), \ u_2 (R+r \ v_1), \ r \ v_2)$. This is clearly a smooth map and $F(S^1 \times S^1) = T$. The inclusions $i : S^1 \times S^1 \hookrightarrow \mathbb R^4$ and $j : T \hookrightarrow \mathbb R^3$ are smooth submanifold embeddings. Hence $F \circ i$ is smooth. Since $F \circ i = j \circ f$, we conclude that $f$ is smooth.
Now consider the map $s : \mathbb R^2 \to S^1 \times S^1, s(x,y) = (\cos x, \sin x, \cos y, \sin y)$. It is smooth since $S = i \circ s$ is smooth. Moreover, $F \circ S = F \circ i \circ s = j \circ f  \circ s$. The Jacobian of $F \circ S$ at $(x,y)$ is
$$\left( \begin{array}{rrrr}
-\sin x(R+r\cos y) & \cos x(R-r\sin y) \\
\cos x(R+r\cos y)  & \sin x(R-r\sin y)   \\
0 & r\cos y  \\
\end{array}\right) $$
It has rank $2$ (compute the determinant of the first two rows and note that both $R+r\cos y,R-r\sin y$ are non-zero). Hence $D_{(x,y)}(F \circ S) = D_{(x,y)}(j \circ f \circ s) = D_{f(s(x,y))}j \circ D_{s(x,y)}f \circ D_{(x,y)}s$ has rank $2$. This implies that $D_{s(x,y)}f$ has rank $2$. Since $s$ is surjective, we see that each $D_pf$ has rank $2$, i.e. is a linear isomorphism.
