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The following is from page 14 of the book heat kernels and Dirac operators

Definition: let $\pi: \mathcal{E} \rightarrow M $ be a smooth map from a manifold $\mathcal{E}$ to a manifold $M$. We say that $(\mathcal{E},\pi)$ is a fiber bundle with typical fiber $E$ over$M$ of there is a covering of $M$ by open sets $U_i$ and diffeomorphisms $\phi_i : \pi^{-1} (U_i) \rightarrow U_i \times E,$ such that $\pi: \pi^{-1}(U_i) \rightarrow U_i$ is the decomposition of $\phi_i$ with projection onto the first factor $U_i$ in $U_i \times E$. The space $\mathcal{E}$ is called the total space of the fiber bundle, and $M$ is called the base.

It follows from the definition that $\pi^{-1}(x)$ is diffeomorphic to $E$ for all $x \in M$.

My goal is to prove the last statement.

Let $x$ be in some $U_i$ in $M$. Since $\pi_i$ is the decomposition of $\phi_i$ with projection onto the first factor $U_i$ in $U_i \times E$. Then $\phi_i$ restricted to $\pi^{-1}(x)$ has an image in $\lbrace x \rbrace\times E$. Since $\phi$ is injective then its restriction to $\pi^{-1}(x)$ is also injective. But I'm not able to conclude why it's surjective ?

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1 Answer 1

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Here's a hint. The map $\phi_i$ is a diffeomorphism, so it's inverse $\phi_i^{-1} : U_i \times E \to \pi^{-1}(U_i)$ is also a diffeomorphism.

Now ask yourself: What is the image of $\{x\} \times E$ under the diffeomorphism $\phi_i^{-1}$?

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  • $\begingroup$ Got it, thanks for the hint @Lee Mosher. $\endgroup$
    – Soumia
    Feb 11, 2022 at 17:05

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