Uniqueness (up to homeomorphism) of the fiber of a locally trivial fiber bundle with connected base

Question

Let $\pi:E\rightarrow M$ be a locally trivial fiber bundle, i.e. for every $x\in M$ there is a neighbourhood of $U_x\ni x$, $U_x\subset M$, a space $F_x$, and a homeomorphism $\phi_x:\pi^{-1}(U_x)\rightarrow U_x\times F_x$ such that $\pi=\pi_1\circ \phi_x$, $\ \pi_1:U_x\times F_x\rightarrow U_x$. That would give us a local trivialization $\{U_i, \phi_i\}$ and a cover $\{U_i\}$.

I want to show that for connected component of $M$ (let's just assume $M$ is connected) we can choose one $F$ and every $F_x$ will be homeomorphic to it.

1. Is it true that $\phi_b:\pi^{-1}(b)\rightarrow \{b\}\times F_b$ is a homeomorphism (as well as $\tilde\phi_b:\pi^{-1}(b)\rightarrow F_b,\ \tilde\phi_b=\pi_2\circ\phi_b$)? My assumption is that we can consider it as a restriction of $\phi_b:\pi^{-1}(U_b)\rightarrow U_b\times F_b$ with $\phi_b(\pi^{-1}(b))=\{b\}\times {F_b}$. But how can I show it in a more formal way?

2. My next step is to show that on intersection of $U_x\cap U_y\ni z$ the following map $\phi_z|_{U_x\cap U_y}\circ\phi^{-1}_z|_{U_x\cap U_y}: \{z\}\times F_x\rightarrow \{z\}\times F_y$ is a homeomorphism, hence $F_x\cong F_y$.

3. Finally I want to use that for connected spaces the following is true: for every collection $\{U_\alpha\}$ of open subsets of $M$ with $M=\cup U_\alpha$ and every two points $x,y\in M$, there are finitely many $U_1,U_2,...,U_n∈\{U_\alpha\}$ such that $x\in U_1, U_i\cap U_{i+1}\ne \emptyset$ for all $1\leq i< n$, and $y\in U_n$. But is it the only way how we can prove it? Is it crucial to use finiteness here?

Answer 1

1. Since you want to be careful about continuity of restrictions, it is good to be explicit about these restrictions in the notation. But yes, $\phi_b | \pi^{-1}(b)$ is continuous because $\phi_b$ is continuous. Similarly, $(\phi_b | \pi^{-1}(b))^{-1} = \phi_b^{-1} | (\{b\} \times F_b)$ is continuous because $\phi_b^{-1}$ is. It is a property of the product topology that $\pi_2 : U_b \times F_b \to F_b$, when restricted to $\{b\} \times F_b$ is a homeomorphism, the inverse $F_b \to \{b\} \times F_b$ is given by $f \mapsto (b, f)$, which is continuous because each coordinate is continuous.

If you want to prove (2), you should actually generalize (1) to show that, for any $b' \in U_b$, the map $\pi^{-1}(b') \xrightarrow{\phi_b} \{b'\} \times F_b \xrightarrow{\pi_2} F_b$ is a homeomorphism.

2. This is the composition of the homeomorphism from (1) applied to $z \in U_y$ and the inverse of the homeomorphism from (1) applied to $z \in U_x$.

3. I think it's easier to show that the set of $x$ for which $F_x$ is homeomorphic to your fixed $F$ is both open and closed. Both of these follow from the (generalized version of) 1, since $F_b' \cong F_b$ for any $b' \in U_b$ so $F_b'$ is homeomorphic to $F$ iff $F_b$ is.