Working out an example in Hatcher vol. $2$: Pullbacks of the Möbius Bundle I'm working out the examples made by Hatcher to shows some pullbacks (definition here for clarity) and this ("simplified" version with $n=2$ or $n=3$) gave me an hard time: 
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Notation: $E$ is the Möbius Bundle $\left( \dfrac{S^1 \times \mathbb{R}}{\sim} , S^1 , \rho \right)$ where the identification is the usual $(z,t) \sim (-z,-t)$ and the map is $\ \rho : E \to S^1 \ \ \ (a,t) \mapsto a^2$. 
The symbol $f^*(E)$ stands for the pullback bundle of $E$ via the map $f$.
With $[a,v]$ I'll denote the equivalence class of $(a,v) \in S^1 \times \mathbb{R}$
$$\times \times \times \times $$

Given $$ f : S^1 \to S^1 \ \ \ \ \ \ z \mapsto z^2$$ prove that $f^*(E) \approx S^1 \times \mathbb{R} $

My attempt
Considering this diagram I only need to define a function $f'$ such that it is continuous and restricted to the fiber $$f'_{|(S^1 \times \mathbb{R})_z} : (S^1 \times \mathbb{R})_z \to E_{z^2}$$ is an isomorphism of linear space. 

So I defined $f'((z,v)) = [z,v]$. If I restrict the  domain to $(S^1 \times \mathbb{R})_z$ the image is $E_{z^2}$, it's continuous and an isomorphism (in each fiber) because the equivalence relation is between fibers, and not "inside" a fiber, so in fact it is like the identity over each fiber. So by the uniqueness property of the pullback I conclude that the pullback is trivial.

Is it correct? I'm not very convinced about my reasoning (the result is true) just because I can't see where it doesn't work if I substitute the map $f : z \mapsto z^2$ with $\tilde{f}: z \to z^3$. I think my reasoning would go the same (mutatis mutandis) but the answer in this case is that $\tilde{f}^*(E) \approx E$
So where is the error (if any) in my reasoning? and Can someone shows formally why the situation is completely different if I change $f$ with $\tilde{f}$? 
NB: I wrote formally because intuitively I know why they are different.
NB2: It seems that my problem is why $z^{3/2}$ is not well defined. I can't think of a reason, just because the same reason would apply to $z^{4/2}$ too, and I think in the latter case would be not true. So the answer must contain explicitly the reason of the not-well-definition of the case with $n$ odd.
NB3: If someone look the edit history could imagine that I create an entirely different question, but I only edited it according to my progress in the resolution of this exercise. I think the problem is there so I stressed it in the question
 A: In order to prove that $\tilde f^*(E)$ is isomorphic to $S^1 \times \mathbb{R}$ you need to construct a map $\tilde f' : S^1 \times \mathbb{R} \to E$ having the property that its restriction to the fiber $(S^1 \times \mathbb{R})_z$ is a linear isomorphism between $(S^1 \times \mathbb{R})_z$ and $E_{z^3}$. For this purpose you cannot use the exact same formula $\tilde f'(z,v)=[z,v]$, because under that formula the image of the fiber $(S^1 \times \mathbb{R})_z$ is $E_{z^2}$, not $E_{z^3}$. So you might decide instead to use the formula
$$\tilde f'(z,v) = [z^{3/2},v]
$$
But now you run into an ambiguity: $z^{3/2}$ is not well-defined, there are two choices. That would be OK, as long as $[z^{3/2},v]$ is well-defined. But its not, because if $w, w'$ are the two choices for $z^{3/2}$ then the two points $[w,v],[w',v] \in E$ are not equal.  
The same happens when replacing $3$ by any odd integer, but when replacing by an even integer $n$ then $z^{n/2}$ is well-defined and no ambiguity arises.
Further remarks, addressing comments:
The square root function on the complex plane is not well-defined: every complex number except zero has two square roots. More generally, for any odd integer $k$, every complex number $z \ne 0$ has two "$k/2$-powers", namely the two square roots of $z^k$. So if you decided to use the formula $\tilde f'(z,v) = [z^{3/2},v]$, you would be forced to check that the expression $[z^{3/2},v]$ is well-defined independent of the choice of a square root of $z^3$. But it is NOT well-defined, as one can easily verify. So this attempt at a proof fails (as it must, because the pullback bundle is not the trivial bundle when $k$ is odd).
On the other hand, if $k$ is even, this problem does not arise, because it is entirely unnecesary to consider "two square roots of $z^k$". Instead you divide $k$ by $2$ and get an integer $k/2$, plug that integer into the expression $z^{k/2}$ and compute the power, then plug that result into the formula $\tilde f'(z,v) = [z^{k/2},v]$ which works like a charm, proving that the pullback bundle is trivial.
