The bundle vector $f^\ast(\xi)$ for Moebius over $S^1$ Take the Moebius band like a vector bundle $\xi$ over the circle $S^1$ and the functions $f_n(z)=z^n$ then my question is: how describe the vector bundle define for the pullback $f^\ast(\xi)$ for every $n\in\mathbb{N}$. I suspect that for n = 2 I get the trivial vector bundle because for antipodal points I have the same fiber defined from the roots of each element. Anyway I can not argue formally and guess it's the same for odds numbers but I do not know what to do with even numbers.
 A: Pullbacks might look not very handy in the beginning. So it is important to make your statements precise and formal, to learn how to deal with them. First we want to consider the case $n=2$.
If we have $M \stackrel g\to S^1$ the twisted line bundle, then we can restrict $g$ to the unit sphere bundle $S$ (after choosing a metric). This will be a two fold covering, hence $S\cong S^1$. Let $i$ be the inclusion. We have the following commutative diagram and the universal property of a pullback gives us a unique section $k$ of the vector bundle $f^*M$. This section is nowhere zero, since $S^1 \stackrel k\to f^*M\to M = i$ and $i$ is nowhere zero and $f^*M \to M$ is a bundle map, hence preserves the zero section. A line bundle with a nowhere zero section is trivial, hence $f^*M \cong S^1\times \mathbb R$.
Now ask yourself again: where did the magic happen? The answer is: in the commutativity of the diagram with $z^2$, i.e. $gi=z^2$.

Lastly you want to show very similarly that for odd degrees you get the twisted bundle and for even degrees you get a trivial bundle. I want to leave that for you to practice. The procedure is very similar to above, note that $2\mathbb Z \supset 4 \mathbb Z \supset \cdots$ and apply what you know about covering spaces.
A: To classify line bundles over $\mathbb S^1$ one needs some tool: cohomology is a good one. A line bundle $E$ is described by a trivialization atlas $\{\varphi_i:U_i\to E\}$, which in turn reduces to its transition functions $g_{ij}:U_i\cap U_j\to\mathbb R$ (since the fibers $E_x$ are lines, automorphism of them are simply multiplication times a non-zero escalar $g_{ij}(x)$). This  gives a $1$-cocycle $(g_{ij})$ in $H^1(\mathbb S,\mathcal C^*)$, where $\mathcal C^*$ stands for non-zero continuous functions. But using the real exponential, what matters of such a cycle are the signs, which gives a cocycle $\big(\text{sign}(g_{ij})\big)$ in $H^1(\mathbb S^1,\mathbb Z/2)$. 
This is a rough and drafty description of the classical isomorphism $\sigma:V^1(\mathbb S^1)\equiv H^1(\mathbb S^1)$ between isomorphism classes of line bundles and the first mod $2$ cohomology group. 
Since $H^1(\mathbb S^1,\mathbb Z/2)=\mathbb Z/2$ we get two isomorphism classes: trivial and Moebius. We denote them $\omicron$ and $\xi$ (your Moebius), so that $\sigma(\omicron)=0,\sigma(\xi)=1$.
Next we have naturality: this isomorphism respects pullbacks. A continuous map $f:\mathbb S^1\to\mathbb S^1$ defines two homomorphisms 
$f^*\!:V^1(\mathbb S^1)\to V^1(\mathbb S^1)$ and $\ f^*\!:H^1(\mathbb S^1,\mathbb Z/2)\to H^1(\mathbb S^1,\mathbb Z/2)$. It holds  $f^*\circ\sigma=\sigma\circ f^*$.
(This is quite straightforward from the construction.)
Finally, $f^*$ has a simple description in cohomology. Since $H^1(\mathbb S^1,\mathbb Z/2)=\mathbb Z/2$, the homomorphism $f^*:\mathbb Z/2\to\mathbb Z/2$ is multiplication times $d_2=0$ or $d_2=1$. This $d_2=\deg_2(f)$ is called degree mod $2$ and coincides with the parity of the generic fiber $f^{-1}(a), \ a\in\mathbb S^1$.
All in all we have the following, 
$$
\sigma(f^*\xi)=f^*\sigma(\xi)=d_2
=\begin{cases}0&\text{if $\#f^{-1}$ is even},\\
1&\text{if $\#f^{-1}$ is odd}.
\end{cases}
$$
In other words, $f^*\xi$ is trivial iff $f$ is even-to-one. In our case $f(z)=z^n$ this means that $f^*\xi$ is trivial when $n$ is even and the Moebius band when $n$ is odd.
This may sound a bit elaborated, but hits several important notions and relations among typical objects in algebraic topology, and might arise some curiosity to learn them further. In any case I hope it's of some help.
