For $0\lt r \lt R$ we look at $f: \Bbb R^2 \to \Bbb R^3 $, $$f(\phi, \psi ) = ((R+r\cos\varphi ) \cos \psi, (R+r\cos \varphi) \sin \psi, r \sin \varphi)$$
I've already shown that $f$ is an immersion, by showing that it's Jacobian matrix hat full column-rank. Now I want to show that the image of $f$, called $T\subset \Bbb R^3$, is a submanifold of $\Bbb R^3$. In our lecture we've shown that the image of a $\mathcal C^k$ immersion is locally a $\mathcal C ^k$ submanifold, maybe this helps.
I don't really know how to widen the result that the image is locally a submanifold to it just beeing a submanifold for all it's points. Also maybe someone has some general tipps on showing that sets are submanifolds, since I'm not quite sure on the subject yet.
Any tipps or ideas are greatly appreciated. Thanks in advance!
To show that $L \subset \mathbb{R}^3$ is a $\mathcal{C}^k$- surface, you must show that for for each $p \in L$, there exists an open set $U \subset \mathbb{R}^3$ and a homeomorphism $\phi_p: U \cap L \to \mathbb{R}^2$. The $\mathcal{C}^k$ structure comes from the additional requirement that any two such homeomorphisms,
$$\phi_i: V \cap L \to \mathbb{R}^2 \ \ \textrm{ and} \ \ \phi_j: U \cap L \to \mathbb{R}^2$$
satisfy $\phi_{i} \circ \phi_j^{-1}\Bigr|_{\phi_j(U \cap V \cap L)}$ is $\mathcal{C}^k$. Maps like these are called $\textbf{local charts}$. Your $f$ actually gives the coordinates of the torus.
You've shown that $\textbf{J}(f)$ has full-rank for all $p=(\phi, \psi)$, which by the Inverse Function Theorem implies $f$ is a local diffeomorphism i.e there exists $ p \ni U$ such that $f: U \to f(U):=V'$ is a diffeomorphism. Hence, we have $f^{-1}: V' \to U $. Since $V'$ is an open subset of $\textbf{Im}(f):=L$ i.e if we assume it inherits the subspace topology, then $V' = U' \cap L$ for some $U'$ and so,
$$f^{-1}: U' \cap L \to U$$
We have an actual collection corresponding to each $\alpha \in U$ i.e $\phi_{\alpha}:= f^{-1}_{\alpha}:U'_{\alpha} \cap L \to U_{\alpha}$. We also have $\phi_{\alpha} \circ \phi_{\beta}^{-1} = f^{-1}_{\alpha} \circ f_{\beta} = (f^{-1} \circ f)\Bigr|_{U_{\alpha} \cap U_{\beta} \cap L}=\textbf{id}$, which is smooth i.e $L$ is a smooth surface. The image is diffeomorphic to a cylinder.