# Bijective fiber bundle map is a bundle isomorphism proof attempt

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.

## 1 Answer

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.