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The canonical definition of fibre bundle is the following:

Let $B,X,F$ be three topological spaces and $\pi:X\rightarrow B$ a continuous surjective map; then $(X,F,B,\pi)$ is a fibre bundle on $B$ if for all $b\in B$ exist an open neighbourhood of $U$ of $b$ and a homeomorphism $\phi_U:\pi^{-1}(U)\rightarrow U\times F$ such that $proj_U\circ \phi_U=\pi_{|U}$ (where $proj_U$ is the canonical projection on $U$).

A consequence of the previuous definition is that the set $\pi^{-1}(p)$ is homeomorphic to $F$ for all $p\in U$, but I don't understant why this is true. Lets try to resctrict the function $\phi_U$ to the set $\pi^{-1}(p)$ (under the hypothesis that $p\in U$), so we have: $$proj_U\bigg(\phi_U(\pi^{-1}(p))\bigg)=\pi_{|U}(\pi^{-1}(p)) = p $$

but now we can't multiply both sides for $proj_U^{-1}$ because the function is not injective, and we can't conclude that $\phi_U(\pi^{-1}(p))=proj^{-1}(p)=\{p\}\times F$.

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If $\phi_U$ is a homeomorphism, then $\phi_U$ restricted to $\pi^{-1}(p)$ will also be a homeomorphism. But this is just a map from $\pi^{-1}(p)$ to $p\times F$ and $p\times F$ is homeomorphic to $F$.

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But $\phi_U$ restricted to $\pi^{-1}(p)$ is a homeomorphism between $\pi^{-1}(p)$ and $p\times A$ where $A\subseteq F$. I'm sorry if this observation will be stupid! – Dubious Oct 18 '12 at 18:46
Who does ensures that the function is surjective on $p\times F$? This is the point. – Dubious Oct 18 '12 at 18:54
You need to use the condition $\textrm{proj}_U \circ \phi_U = \pi |_U$. – Zhen Lin Oct 18 '12 at 19:50

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