# Question about fiber bundles

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 ?

## 1 Answer

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 $$\phi_i^{-1}$$?

• Got it, thanks for the hint @Lee Mosher. Feb 11, 2022 at 17:05