I am working through a couple of proofs concerning the Mobius Strip, and I am wondering if the following proof is valid. The claim is:
CLAIM
The Mobius Strip is defined by $M=${$(e^{i\phi},re^{i\frac{\phi}{2}}):\phi\in[0,2\pi],r\in[-1,1]$} $\subset\mathbb{S}^1\times\mathbb{C}$. We define $\pi:M\to\mathbb{S}^1$ by $\pi=\text{pr}_1|_M$. This manifold $M$ is a fiber bundle with general fiber $F=[-1,1]$.
PROOF
We must show that for all $x\in M$, there exists an open nbhd $U$ and diffeomorphism $\theta_U:\pi^{-1}(U)\to U\times F$ such that $\text{pr}_1\circ \theta_U=\pi$. We define $\theta_U:\pi^{-1}(U)\to U\times F$ as the map $x\mapsto (\pi(x),\text{pr}_2|_M(x)\pi(x)^{-\frac{1}{2}})$. The function defined in this way is surjective because $\pi^{-1}(x)=${$(x,r):r\in[-1,1]$} and $\text{pr}_2|_M(x)\pi(x)^{-\frac{1}{2}}=r$. It is injective because $\pi(x)=\pi(y)$ fixes the first coordinate, while $\text{pr}_2|_M(x)=\text{pr}_2|_M(y)$ fixes the second coordinate. Using the (complex) basis {$\pi(x),\text{pr}_2|_M(x)$} as a basis for $\mathbb{S}^1\times\mathbb{C}$, we have that the differential $d\theta_U$:
$d\theta_U(x)=\begin{pmatrix}1&0\\\ -\frac{1}{2}\text{pr}_2|_M(x)\pi(x)^{-\frac{3}{2}}&\pi(x)^{-\frac{1}{2}}\end{pmatrix}$
Which is always nonsingular because $\pi(x)\in\mathbb{S}^1$. So, by the inverse function theorem, we can find some open nbhd $U'\subset M$ of $x$ such that $\theta_U$ has a smooth inverse. This neighborhood can be of the form $U'=U\times[-1,1]=\pi^{-1}(U)$. Thus, given any $x\in M$, there exists some nbhd $U'$ of $x$ and diffeomorphism $\theta_U:\pi^{-1}(U)\to U\times F$ (it is obvious here that $\text{pr}_1\circ \theta_U=\pi$), i.e., every point $x\in M$ has a local trivialization, thus, $M$ is a fiber bundle.
$\blacksquare$
Is this proof correct? Is there any way to make it cleaner / more concise?
Thank you in advance.