Topology of the tangent bundle of a smooth manifold I am having trouble understanding 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 separated 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.
 A: You give $TM$ a topology and a manifold structure as follows. Suppose that $\varphi\colon U\subseteq M\to V\subseteq\mathbb{R}^n$ is a local chart of $M$. Let $x_1,\ldots, x_n$ be the corresponding coordinate functions, i.e., $\varphi(p) =(x_1(p),\ldots, x_n(p))$. Then you get a bijective map $\pi^{-1}(U)\to V\times \mathbb{R}^n$ given by $$\left(p, \sum_{i=1}^n\lambda_i\left.\frac{\partial}{\partial x_i}\right|_p\right)\mapsto (\varphi(p), (\lambda_1,\ldots, \lambda_n)).$$ The topology on $\pi^{-1}(U)$ is defined by pulling back the topology on $V\times \mathbb{R}^n$. Moreover, this map $\pi^{-1}(U)\to V\times\mathbb{R}^n$ is a chart map for the manifold $TM$. You must check, of course, that if you choose different charts $\varphi$, this does not change the topology or the manifold structure.   
A: Take some atlas on $M$, and let $U$ be an element of that atlas. Then $TU=\pi^{-1}(U) \cong U \times \mathbb{R}^n$ as a set, so it inherits a topology. Moreover, all these topologies (for different $U$) are compatible with each other, so together they give you a topology on the total space $TM$.
Note that this is very much like your second idea, except that we don't require open sets to contain entire fibres $\pi^{-1}(x)$ -- just open subsets of them.
A: Maybe a broader perspective would be useful.
Let $F$ and $B$ be topological spaces, with a given open cover $\{U_\alpha\}$ of $B$ and continuous transition functions $\theta_{\alpha\beta}:U_\alpha\cap U_\beta\to \operatorname{Aut}(F)$ on nonempty intersections of the sets in the open cover, which satisfy the conditions that, always, $\theta_{\alpha\beta} = \theta_{\beta\alpha}^{-1}$, $\theta_{\alpha\alpha} = 1$, and $\theta_{\alpha\beta}\theta_{\beta\gamma} = \theta_{\alpha\gamma}$. 
These data are exactly what we need to put together a fiber bundle with fiber $F$ and base $B$.  We do this the following construction.  One consequence of the construction is insight into the topology of the total space of the fiber bundle.
The prospective bundle atlas will be composed of prospective bundle trivializations $\{U_\alpha\times F\}$, indexed by the open cover of $B$.  Now define 
$$E = \bigg(\coprod_\alpha U_\alpha\times F\bigg)/\sim,$$
where $(x,f)\sim (y,g)$ if and only if $x=y\in U_\alpha\cap U_\beta$ and $\theta_{\alpha\beta}(x)f = g$.  (Note that the topology of each component of the disjoint union is just the product topology.)
While it is an exercise (left to you) to check that this is in fact the total space of a fiber bundle, the idea should be clear enough: we have taken the locally trivial neighborhoods and pasted them together with knowledge of how they transform into each other.  We now see that the topology of the total space is just the quotient topology.
This construction encompasses all topological fiber bundles.  To construct smooth fiber bundles, replace "continuous" by "smooth" and "topological space" by "smooth manifold."  Nice examples include (but are certainly not limited to):


*

*the Hopf fibrations $\mathbb{S}^1\to\mathbb{S}^{2n+1}\to\mathbb{C}P^n$, $\mathbb{S}^3\to\mathbb{S}^{4n+3}\to\mathbb{H}P^n$; 

*symmetric and homogeneous spaces, such as the fibration $SO(n)\to SO(n+1)\to\mathbb{S}^n$; 

*Seifert fibered spaces and surface bundles in $3$-manifold theory; and 

*all real and complex vector bundles, such as $TM$, $T^*M$, the exterior bundles $\Lambda^k(M)$, and the tensor bundles $\mathcal{T}^r_s(M)$, for a smooth manifold $M$.


In the particular case of your question, we have $F = \mathbb{R}^n$, transition functions are maps $U_\alpha\cap U_\beta \to Gl(n;\mathbb{R})$, and the topology is given locally by the product topology on $U\alpha\times\mathbb{R}^n$.  The "compatibility" mentioned in the Micah's answer, and the check froggie suggests, are just verification that the topology on $TM$ is the quotient topology in my above definition.
A: 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 $M$? 
