Vector bundle is Manifold

Question

I want to show that the total space of a vector bundle $E \overset{\pi}{\rightarrow} M$ is itself a manifold. It is easy to construct charts by just composing the bundle maps $\pi^{-1}(U_{\alpha}) \rightarrow U_{\alpha} \times \mathbb{R}^k$ with the charts for $M$. It now remains to show the Hausdorff property and that $E$ is second countable. Can someone help me with that?

Answer 1

This is a little outside my wheelhouse, so someone please point errors out if there are any. EDIT: I've noticed a few errors in my original answer, with the help of OP's prompt, and my corrections follow this answer: https://math.stackexchange.com/a/213388/

For Hausdorffness, let $x,x' \in E$ be two distinct points. There are two cases. For the first case, if $\pi(x) \neq \pi(x')$, then since $M$ is Hausdorff, we can choose disjoint open sets $U, U'$ with $\pi(x) \in U$, $\pi(x') \in U'$. Then $\pi^{-1}(U)$ and $\pi^{-1}(U')$ are disjoint open sets with $x \in \pi^{-1}(U)$ and $x' \in \pi^{-1}(U')$. Note that for this case it is not necessary to guarantee that $U, U'$ be coordinate domains.

For the second case, suppose $\pi(x) = \pi(x') =p$. Then choose a coordinate domain $U$ in $M$, containing $p$, with a homeomorphism $\phi: \pi^{-1}(U) \to U \times \mathbb{R}^k$. Since $U \times \mathbb{R}^k$ is Hausdorff, take disjoint open sets $V, V' \subset U \times \mathbb{R}^k$ with $\phi(x) \in V$, $\phi(x') \in V'$. Then $x \in \phi^{-1}(V)$ and $x' \in \phi^{-1}(V')$, and these are disjoint open sets in $E$.

For second countability, take a countable base $\mathcal{B}$ for $M$, then for each $U \in \mathcal{B}$, take a countable base for $\pi^{-1}(U) \simeq U \times \mathbb{R}^k$.