TITLE: Topology of the tangent bundle of a smooth manifold
QUESTION [38 upvotes]: I  am having trouble to understand what topology is given to the tangent bundle of a smooth manifold that allows it to be a smooth manifold itself. In my understanding, among other things the topology must be second countable and Hausdorff. The definition of the tangent bundle $TM$ of a smooth manifold $M$ I am using is 
$TM = \bigsqcup_{p\in M} T_pM$,
that is the disjoint union of all $T_pM$ where $T_pM$ is tangent space at $p$ consisting of all derivations at $p$. Since there is no further specification on what topology this space is given I assume we take the natural disjoint union topology. 
However, in that case it seems that $TM$ is not second countable because then every set $(O,p)$ where $O$ is an open subset of $T_pM$ would be open and disjoint from any $(O,q)$ for $q \neq p$. So unless $M$ is countable there would be an uncountable number of disjoint open sets which contradicts second countability.
The only alternative I can think of is using the natural smooth structure of $TM$ as the topology. That is for every open subset $O$ of $M$ the open sets of $TM$ are defined as $\pi^{-1}(O)$ where $\pi$ is the natural projection $TM \rightarrow M$.
But then $TM$ can not be Hausdorff, since any two elements of the same fiber of $\pi$ could not be seperated by open sets.
In conclusion, in both cases $TM$ could not be a manifold, so I must be missing something very obvious. Thus, I would really appreciate it if someone could point out my misconception.

REPLY [15 votes]: I'll expose the construction of the topology on $TM$ which pleases me most. I won't fill in the details but I'll add a reference later which has complete proofs.
Definition: Let $(U, \phi)$ be a local chart of $M$. We define $$\phi_{TM}:\pi^{-1}(U)\longrightarrow \phi(U)\times \mathbb R^n, v\longmapsto ((\phi\circ \pi)(v), d\phi_{\pi(v)}(v)).$$
The map $\phi_{TM}$ is bijective and we will use this to induce a topology on $TM$.
Let $\mathfrak{A}$ be a smooth atlas for $M$. Define: $$\mathscr{T}_{TM}:=\{W\subseteq TM: \phi_{TM}(W\cap \pi^{-1}(U))\ \textrm{is open in}\ \mathbb R^n\times \mathbb R^n\ \forall\ (U, \phi)\in\mathfrak{A}\}.$$
Exercise 1. Show this is a Hausdorff topology and has a countable basis of open sets and all the maps $\phi_{TM}$ are homeomorphisms. 
Just to finish, the smooth structure on $TM$ will be given by the atlas $$\mathfrak{A}_{TM}:=\{(\phi_{TM}, \pi^{-1}(U)): (U, \phi)\in \mathfrak{A}\}.$$
You may check this topology turns the canonical projection $\pi:TM\longrightarrow M$ into a smooth map, this is not hard to see for when using charts you will get a map which is a projection between open sets of $\mathbb R^n$.
Exercise 2. We made a choice of an atlas $\mathfrak{A}$ for defining the topology of $TM$ and its smooth structure. What would happen if you had started with a different smooth atlas on $M$?