6

Let $(E,\pi,M,F)$ is a fibre bundle. The map $\pi : E \rightarrow M$ is the projection map. The smooth manifolds ; $E$ is the total space, $M$ is the base space, and $F$ is a smooth manifold appear in the following local trivialization : For any $p \in M$, there are open subset $U\subset M$ contain $p$ and diffeomorphism

$$\phi : \pi^{-1}(U) \rightarrow U \times F$$

such that $\pi = Proj_U \circ \phi$ on $\pi^{-1}(U)$, where $Proj_U : U \times F \rightarrow U$ is the projection to the first factor.

$\textbf{Problem}$ : If $M$ and $F$ connected, show that $E$ is connected.

I already make such construction, but i'm not really sure and curious about other solution too. Feel free to post new proof or disprove my proof :

I'm having hard time with this, but i end up trying to prove this by using the following theorem from topology (e.g from Willard's book theorem 26.15) : If a topological space $X$ is connected and $\mathscr{U}$ is an open cover for $X$, then any two points can be connected by a simple chain consisting of elements of $\mathscr{U}$.

By local trivialization for each $p \in M$ we have an open subset $U \subset M$ containing $p$ and a diffeomorphism $\phi : \pi^{-1}(U) \rightarrow U \times F$. Let $\{\pi^{-1}(U)\}$ be the open cover for $E$. Because $F$ and $M$ connected, then $U\subset M$ connected, $U \times F$ connected, $\phi^{-1}(U\times F) = \pi^{-1}(U)$ connected. With this open cover, we have a simple chain where each of its elements is connected (implies path-connectedness). By this we can easily make a path connecting $v,w \in E$ by joining the paths from each chain.

Kelvin Lois
  • 7,430

1 Answers1

6

Suppose that $E$ is not connected, there exists a surjective continuous function $f:E\rightarrow \{0,1\}$ write$U=f^{-1}(0), V=f^{-1}(1)$, $E=U\cup V$ where $U$ and $V$ are not empty, let $\pi:E\rightarrow M$ the projection map. Remark that $\pi$ is an open map. This implies that $\pi(E)=M=\pi(U)\cup (V)$ is the union of the open subsets $\pi(U)$ and $\pi(V)$ wich are not disjoint since $M$ is connected. Let $x\in \pi(U)\cap \pi(V)$, there exists $y_1\in U, y_2\in V$ such that $\pi(y_1)=\pi(y_2)=x$, since $\pi^{-1}(x)$ is connected, the restriction of $f$ to $\pi^{-1}(x)$ is constant, but $f(y_1)=0, f(y_2)=1$. Contradiction.

Tsemo Aristide
  • 89,587