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On defining a fibre bundle, it is argued that the projection map $\pi$ requires to be satisfied the condition that there is a homeomorphism $h$ such that the first coordinate coincide.

Definition (Fibre bundle). A fibre bundle structure on a space $E$ is a pair of spaces $F$ and $X$ equipped with a projection map $\pi: E \rightarrow X$ such that there is an open cover $\left\{U_i\right\}$ of $X$ and homeomorphisms $h_i: \pi^{-1}\left(U_i\right) \rightarrow U_i \times F$ coinciding with $\pi$ in the first coordinate. We write: $$ F \longleftrightarrow E \stackrel{\pi}{\longrightarrow} X . $$ In particular, this means that each fibre $F_x=\pi^{-1}(x)$ maps homeomorphically to $\{x\} \times F$. We call $E$ the total space, $X$ the base space, $F$ the fibre and $h_i$ the (local) trivialisations.

But I don't understand how this two spaces $\pi^{-1}\left(U_i\right)$ and $U_i \times F$ are really different?

Eden Zane
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But I don't understand how this two spaces $\pi^{-1}(U_i)$ and $U_i\times F$ are really different?

They are not "really" different in the sense that they are diffeomorphic, meaning "essentially the same".

However, they are not "the same" in the sense that $\pi^{-1}(U_i)\subseteq E$ and $U_i\times F$ is not related to $E$ (Note that $E$ and $F$ are both manifolds).

So the statement that $E$ has a fibre structure means that locally the space $E$ looks like $U_i\times F$, i.e. there are open subsets $\pi^{-1}(U_i)$ that look like/are diffeomorphic to $U_i\times F$.

However, $E$ might not have the global structure $E\cong X\times F$ (not necessarily true). For example, let $E$ be a Moebius strip and let $X$ be a circle on this Moebius strip (going around once). Now for each $x\in X$ draw a line segment on the strip such that the whole strip is nicely divided up into line segments (As in this wikipedia image). Then globally the space $E$ does not look like "a circle"$\times$"a line segment". That would be a cylinder, not a Moebius strip.

student91
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  • what does that mean when they say, the first coordinate must coincide? Does they mean that the point which lies on the base space in two cases must be the same or are there coordinates systems defined on the manifold? – Eden Zane Feb 14 '23 at 11:29
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    They mean the following: There is a canonical projection $p\colon U_i\times F\to U_i$ that sends $(u,f)$ to $u$. Now the function $h_i\colon \pi^{-1}(U_i)\to U_i\times F$ must satisfy $p(h_i(e))=\pi(e)$, i.e. the projection of the element $e\in E$ from $E$ to $X$ (under $\pi$) must coincide with the canonical projection $p\colon U_i\times F\to U_i$ under $h_i$. – student91 Feb 14 '23 at 11:37