Let $f\colon (\phi,\psi) \mapsto \begin{pmatrix} (2+\cos\phi)\cos\psi\\ (2+\cos\phi)\sin\psi\\ \sin\phi \end{pmatrix},\ \phi,\psi\in\mathbb{R}$
be given. I have to show, that $f(\mathbb{R}^2)$ is a submanifold of $\mathbb{R}^3$.
The problem is if I were to choose to points $(\phi,\psi)$ and $(\phi +2\pi,\psi +2\pi)$ then they have the same image and it is therefore impossible to find a homeomorphic map from $f(\mathbb{R}^2)$ to $\mathbb{R}^2$.
I cannot find my mistake (which I assume is there, because the exercise is from a book).
It is impossible IF you're trying to use $f$, restricted to some set in $\mathbb{R}^2$, as your only chart. Can you think of another function, similar to $f$, which covers the set $f(\mathbb{R}^2)$ in a different way? One good example to keep in the back pocket is $S^1$. We see that $t\mapsto(\cos(t),\sin(t))$ wraps around the circle several times, depending on how we choose the parameter domain $t\in[a,b]$. As you point out, we don't want this wrapping because then our chart won't be a homeomorphism. Okay, so lets restrict this chart a bit. Let's use the chart $t\mapsto(\cos(t),\sin(t))$ with parameter domain $(0,2\pi)$. Great! Not too hard to see that this is a homeomorphism onto its image. Problem is, you now notice that we've missed the point $(1,0)$ in the plane, right? Well, we just use a new chart! So to cover $(1,0)$, use the chart $t\mapsto(\cos(t),\sin(t))$ with parameter domain, say, $(-\pi/2,\pi/2)$. It is a little bit weird that our two charts have the same "formula", but morally there's nothing wrong with that.