Let us consider the Mobius Bundle.
- Can Someone Explain the part "There is no natural unique homeomorphism of $Y_x$with $Y$. However there are two such which differ by the map $g$ of $Y$ on itself obtained by reflecting in its midpoint. In this case the group $G$ is the cyclic group of order 2 generated by $g$."
2.If we consider the cylinder over the circle, that is the Space $L \times Y$ without the twist (i.e., identifying the two sides identically without a twist). I know that then we will get a trivial bundle. And the group G will be just identity. But, It appears to me for this case also the group G will be a cyclic group of order 2. Can someone explain it in a clearer way?
Notation: $Y_x=p^{-1}(x)$; $G$= Group of Homeomorphism of $Y$.
The idea underlying these questions is the structure group of the fiber bundle, and the issues raised here are related to the fact that the structure group is somehow specified by other structures and not determined by the topology of the bundle see this question for a general discussion of this.
For both the Mobius band and the cylinder, it is possible to find an open cover $\{U_i\}$ of the circle $X$ such that for each set $U_i$, you can identify $p^{-1}(U_i)$ with $Y \times U_i$. There are many ways of choosing these identifications, and you can try to do so using "as few homeomorphisms of $Y$ as possible." In other words, when $U_i\cap U_j \neq \emptyset$, there will be points in $Y\times U_i$ and $Y\times U_j$ that are identified. You can try to require that only points of the form $(y,x) \in Y\times U_i$ and $(y,x) \in Y\times U_j$ are identified (i.e. $y$ is the same element of $Y$ for each), and doing so will yield an identification of the whole space with $Y\times X$. It is easy to see that this is possible for the cylinder, and the result gives a homeomorphism of each $Y_x$ with $Y$.
But this is not possible for the Mobius band; it has to "flip" somewhere. So even though the identification of $p^{-1}(U_i)$ with $Y \times U_i$ identifies each $Y_x$ with $Y$ for $x \in U_i$, if $x \in U_i \cap U_j$, the identifications of $p^{-1}(U_i)$ with $Y\times U_i$ and $p^{-1}(U_j)$ with $Y\times U_j$ may identify $Y_x$ with $Y$ in different ways. This is what is meant by 1.
In this latter case, you identify points of the form $(y,x) \in Y\times U_i$ with $(\sigma(y),x) \in Y\times U_j$ for homeomorphisms $\sigma:Y\to Y$, and you can try to use "as few $\sigma$'s as possible." Note that the $\sigma$'s are called transition functions. For the Mobius band, it is possible to do this with only two homeomorphisms: the identity, and the flip through the midpoint, so you can restrict the transition functions to $\mathbb{Z}/2$. As you mention in 2., you can certainly do this with the cylinder using transition functions in $\mathbb{Z}/2$, but the point is that it is possible to restrict them to just the trivial homeomorphism. Note that if, for the cylinder, you used $\sigma$'s that flipped $Y$ through the midpoint, you would be ignoring the orientation of the cylinder. This illustrates how restricting the structure group is related to finer structures on the bundle.
For any fiber bundle, the $\sigma$'s described above will range over the homeomorphisms $Y\to Y$, but this is often an extremely large group. Restricting this to a smaller group gives a reduction of the structure group (and allows you to control/understand the structure of the bundle much better). If the fibers are all homeomorphic to a vector space $V$, then you can use arbitrary homeomorphisms $\sigma$, or you can insist that the maps are linear, and your structure group reduces from $HOMEO(V)$ to $GL(V)$. As described in the link above, putting an inner product on each fiber then means you can insist on orthogonal maps, and the structure group reduces to $O(V)$.