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We are working in the continuous category.

Definition 1: A bundle map between two fiber bundles $\pi_1:E_1\rightarrow B$, $\pi_2:E_2\rightarrow B$ is a continuous map $f:E_1\rightarrow E_2$ such that $\pi_2\circ f = \pi_1$

Definition 2: A bundle map $f:E_1\rightarrow E_2$ between two fiber bundles $\pi_1:E_1\rightarrow B$, $\pi_2:E_2\rightarrow B$ over the same base space $B$ is said to be a bundle isomorphism if $f$ is a homeomorphism.

Proposition: If $f:E_1\rightarrow E_2$ is a bijective fiber bundle map covering the same base space $B$, then $f$ is a isomorphism.

My attempt: Let $f^{-1}:E_2\rightarrow E_1$ denote the inverse of $f$. Since $\pi_2\circ f= \pi_1$, it follows that that $\pi_1\circ f^{-1}=\pi_2$.

We show that $f^{-1}$ is locally continuous.

Let $e_2\in E_2$. Consider $\pi_2(e_2)\in B$. Choose a neighborhood $\tilde{U}$ of $\pi_2(e_2)$ and local trivializations

$\phi_2:\pi_2^{-1}(\tilde{U})\rightarrow \tilde{U}\times F_2$

$\phi_1:\pi_1^{-1}(\tilde{U})\rightarrow \tilde{U}\times F_1$

Where $F_1,F_2$ denotes fibers of the bundles.

Hence,

(1). $\pi_{\tilde{U}}\circ \phi_1= \pi_1$

(2). $\pi_{\tilde{U}}\circ \phi_2 = \pi_2$.

where $\pi_{\tilde{U}}$ denotes projection onto $\tilde{U}$.

Since $f$ is a bijective bundle map covering $B$, we have $\pi_2=\pi_1\circ f^{-1}$.

Hence locally, $\pi_{\tilde{U}}\circ \phi_2 = \pi_{\tilde{U}}\circ \phi_1 \circ f^{-1}$ .

However, I am not sure how to verify that $f^{-1}$ is continuous.

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1 Answer 1

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This has no reason to be true. E.g. $B=\{0\}$ and $E_i=B\times F_i$ where $F_1,F_2$ are the same set $F$ but equipped with different topologies, such that ${\rm id}_F:F_1\to F_2$ is continuous but ${\rm id}_F:F_2\to F_1$ is not. Then, $f:={\rm id}_{B\times F}:E_1\to E_2$ is a bijective fiber bundle map but not an homeomorphism.

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