1

I'm getting started with fiber bundles and I have a doubt on the definition. I'll state just the part of the definition I'm in doubt: a fiber bundle is $(E,B,\pi,F,G)$ where $E,B,F$ are topological spaces, $G$ a topological group of homeomorphisms of $F$, $\pi : E\to B$ is a surjective continuous map and there is a open cover $\{U_i\}$ of $B$ such that

  1. For each $U_i$ there's a homeomorphism $\varphi_{i} : \pi^{-1}(U_i)\to U_i\times F$ of the following form $\varphi_i(a) = (\pi(a), \psi_i(a))$. Let $x\in U_i$ we denote $\psi_i | \pi^{-1}(x)$ by $\psi_{i, x}$ and we can see that this is a homeomorphism between $\pi^{-1}(x)$ and $F$.

And then comes other conditions. Now, my doubt there is: how can we see that the mapping $\psi_{i, x}$ is a homeomoprhism? I know that $\psi_{i} = \pi_2 \circ \varphi_i$ where $\pi_2$ is the projection onto the second factor. Both $\varphi_i$ and $\pi_2$ are continuous and surjective, hence $\psi_i$ is continuous and surjective and so is it's restriction $\psi_{i, x}$.

Now, I was failing to see where injectivity comes from. I've tried like this: Let $\psi_{i, x}(a) = \psi_{i,x}(b)$, then we already know that $\psi_i(a)=\psi_i(b)$. Since both $a,b\in \pi^{-1}(x)$ we have that $\pi(a)=\pi(b)=x$ and so, this implies that $\varphi_i(a)=\varphi_i(b)$ and since $\varphi_i$ is homeomorphism this implies $a=b$.

Now the continuity of the inverse I can't see where it comes from. I've tried to show the inverse to be continuous, but I couldn't. How can I show this inverse is continuous?

Gold
  • 26,547
  • 1
    Intuitively, $\psi_{i,x}$ is the restriction of $\phi_i$ to $\pi^{-1}(x)$. The first point $\pi(a) = x$ stays the same as you move around along the preimage over $x$. (Btw, your injectivity proof is perfectly fine.) – Braindead Feb 20 '14 at 13:08
  • Hint: Use $p_2: U \times F \to F$, the projection into the second factor, and $i_x: F \to U\times F$, the inclusion of $F$ as $(x, F)$ in $U\times F$. Note that both of these maps are continuous. – Braindead Feb 20 '14 at 13:10

1 Answers1

0

As a note, observe that this can be also be seen immediately by using the fact that restrictions of homeomorphisms are homeomorphisms. Indeed in our case the subspace topologies in the domain and codomain are respected and $\left.\varphi_i\right|_{\pi^{-1}(x)}:\pi^{-1}(x)\to \{x\}\times F$ as we do all this over $B$.

Looking at the comments by Braindead, what I write above upgrades what he claims to be intuitive to rigorous, for all its worth.

Alp Uzman
  • 10,742