What is difference between inverse of projection map in the domain of homomorphism in defining a fibre bundle and while defining it's cross section? While defining a fibre bundle we talk of homeomorphisms $\phi_{\alpha}: \pi^{-1}(U_{\alpha})\to U_\alpha \times F$ where $\pi:E\to B$ and $U_\alpha$ is an open subset of $B$.
Again we define the section of a bundle as $\pi(\sigma(x))=x,\hspace{0.04cm} \forall x\in B$ which is basically the right inverse of $\pi$.
Is this different from the domain of $\phi_{\alpha}$?
An example shows it really is---I don't understand why it is so from the definition itself. E.g. A global section of a Mobius strip can be a wavy line through the strip, but $\pi^{-1}(U_{\alpha})$ where $U_{\alpha}$ is an arc of the circle $S_1$ is a twisted rectangular shaped strip cut out from the whole band. So these are different.
Is the difference due to the fact that sections has to be injective which the rectangular shaped strip is not?
Note that I am wanting to know the general scenario, not only about the Mobius strip.
 A: I might be misinterpreting what you wrote, but I think what's happening is that you're mixing up two different meanings of the symbol $\pi^{-1}$. So let me try to explain those first in general terms.
Suppose $f\colon X\to Y$ is a map (i.e., a function). (For this discussion, we can just consider $X$ and $Y$ to be arbitrary sets and $f$ to be an arbitrary map between them.) If $f$ happens to be bijective, then it has an inverse map, denoted by $f^{-1}:Y\to X$, which satisfies $f\circ f^{-1} = \text{Id}_Y$ and $f^{-1}\circ f = \text{Id}_X$. Note that this notation only makes sense when $f$ is known to be bijective.
On the other hand, we also use the notation $f^{-1}$ in the following way, regardless of whether $f$ is bijective or not: Given a subset $S\subseteq Y$, the notation $f^{-1}(S)$ denotes the preimage (sometimes called the inverse image) of the set $S$ under $f$: this is the subset of $X$ consisting of all points that are mapped into $S$ by $f$, or more formally
$$
f^{-1}(S) = \{x\in X: f(x)\in S\}.
$$
It would probably have been better to use a different notation for the preimage (and a few authors do so), but this notation is so well established that we all just have to get used to it.
The differences between these concepts are profound, and you must keep them separate:

*

*The inverse map $f^{-1}$ is a map, and makes sense only when $f$ is bijective.

*On the other hand, the preimage $f^{-1}(S)$ is a set, and makes sense for any map whatsoever.

In order to determine which meaning of $f^{-1}$ is intended, look at what it's being applied to: If $f^{-1}$ appears by itself or applied to an element of $Y$ (the codomain of $f$), then it means the inverse map (and $f$ had better be bijective). If $f^{-1}$ is applied to a set, on the other hand, it means the preimage.
Now in the situation you asked about, the notation $\pi^{-1}(U_\alpha)$ refers to a subset of $E$ (because $U_\alpha$ is a subset of $B$). On the other hand, if $\sigma:B\to E$ is a section of $E$, then it is, as you noted, a right inverse for $\pi$. However, this is not the same as an inverse map -- in fact, $\pi$ is almost never bijective (the only exception is when the fibers of $E$ are single points), and so it does not have an inverse map. One would never use the notation $\pi^{-1}$ for a map that is merely a right inverse.
So in short: $\pi^{-1}(U_\alpha)$ is a certain subset of $E$, while $\sigma$ is a map from $B$ to $E$. They are very different animals, and just about the only thing they have in common is that their descriptions might both contain the word "inverse" (but with different meanings).
