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We say a covering space $(E,p)\to X$ is trivializable, if there is a space $F$ (with the discrete topology), and a homeomorphism between $\varphi:E\to X\times F$ such that $p=pr_1\circ \varphi$, where $pr_1:X\times F\to X$ is the projection. Suppose the following proposition holds:

If $B=U\cup V$, both $U$ and $V$ are connected open subsets, and $U\cap V$ is connected. Let $(E,p)$ be a covering space of $B$ such that both $p^{-1}(U)\to U,p^{-1}(V)\to V$ are trivializable, then $(E,p)$ is trivializable.

Show that, if $B\subset\mathbb R$ is an interval, then any covering space of $B$ is trivializable.

The attempt:

We first assert that, if $(E,p)$ is a covering space of $B$, and $B=B_1\cup B_2$, both $B_1,B_2$ are closed, $B_1\cap B_2=\{b\}$, and both $p^{-1}(B_i)\to B_i$ are trivializable, then $(E,p)$ is trivializable.

Let $\varphi_i:p^{-1}(B_i)\to B_i\times F_i$ be the homeomorphisms. Since $B_1\cap B_2=\{b\}$, there is a bijection $\psi:F_2\to F_1$. We define a map $$f:B\times F_2\to E,\;(u,x)\mapsto\begin{cases} \varphi_1^{-1}(u,\psi(x)),&\text{if}\; u\in B_1;\\ \varphi_2^{-1}(u,x),&\text{if}\; u\in B_2. \end{cases}$$ It is not difficult to see $f$ is bijective. We have $f$ is continuous by the pasting lemma and $f^{-1}$ is continuous since $f$ is open. Therefore, $(E,p)$ is trivializable.

For the case that $B$ is a closed and bounded interval $[a,b]$. $\forall x\in B$, there is a neighborhood $U\ni x$ such that $p^{-1}(U)\to U$ is trivializable, then $B$ can be covered by open subsets having this property. By the lebesgue number lemma, we can choose $a=t_0<t_1<\cdots<t_n=b$ such that $p^{-1}([t_i,t_{i+1}])\to [t_i,t_{i+1}]$ is trivializable, for $i=0,1,\cdots,n-1$. Then we have $p:E\to B$ is trivializable by the assertion above.

But how to show it for other cases?

Paul Frost
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ALe0
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  • I can show this propostion for the case that $X$ is a closed bounded interval. I will edit it right away. – ALe0 May 26 '21 at 07:05
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    If you can show it for one case you should edit your question to indicate that (i.e. provide context) and use that to explain why you can't extend it to an arbitrary interval. Without showing that, no-one knows what you understand, or where your difficulty lies, and so cannot write a good answer for you. Also, your question at the moment will be closed for lack of context as it's little more than a demand for someone to solve it for you. – postmortes May 26 '21 at 07:07
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    That's fantastic, a great improvement to the question! Thank-you :) – postmortes May 26 '21 at 08:48

1 Answers1

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Let us first show that it is true for $B = [0, \infty)$.

Define $I_n = [n,n+1]$. You know that each $p_n : p^{-1}(I_n)\to I_n$ has a trivializing homeomorphism $\phi_n : p^{-1}(I_n) \to I_n \times F_n$. This gives homeomorphims $h_n : F_n \to F_{n+1}$ determined by $\phi_{n+1}((\phi_n)^{-1}(x,n)) = (h_n(x),n+1)$ and trivializing homeomorphisms $$\psi_n = h_1^{-1}\dots h_{n-1}^{-1}\phi_n : p^{-1}(I_n) \to I_n \times F_1 .$$ By definition $\psi_n$ and $\psi_{n+1}$ agree on the fiber $p^{-1}(n) = p^{-1}(I_n) \cap p^{-1}(I_{n+1})$. Hence $$\psi : E \to [0, \infty) \times F_1, \psi(e) = \psi_n(e) \text{ for } e \in p^{-1}(I_n)$$ is a trivializing homeomorphism.

Similarly we can treat $B = \mathbb R$.

Each non-compact interval $B$ admits a homeomorphism $H : B \to B'$ to one of $B' = [0, \infty), \mathbb R$. Clearly $p' = Hp : E \to B'$ is a covering. We know that there exists a trivializing homeomorphism $\phi' : E \to B' \times F$. But then $$\phi = H^{-1}\phi' : E \to B \times F$$ is also a trivializing homeomorphism.

Remark:

A much more general result can be found in Are contractible open subsets always evenly covered?

Paul Frost
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