The tangent bundle is the set of all tangent vectors on a manifold. Contrast this with the tangent space at a point, which is just the tangent vectors based at that point. Thus the tangent bundle is the union of all tangent spaces. The tangent bundle inherits a manifold structure and this allows us to define the notion of vector field.
Let \(M\) be a \(C^{\infty}\) manifold of dimension \(n\). The tangent bundle is the set
\begin{align*} TM &= \bigcup_{x\in M} T_xM \\ &= \left\{[c] : c \in C^1\big((-\epsilon, \epsilon) \to M\big)\right\}. \end{align*}Here, recall that \([c]\) denotes the equivalence class of the \(C^1\) curve \(c\) where \(c \sim d\) if \(c(0) = d(0)\) and \((\varphi \circ c)'(0) = (\varphi \circ d)'(0)\) for every chart \(\varphi\) with \(c(0)\) in it's domain.
Show that there is a bijection \(\mathbb{R}^n \times \mathbb{R}^n \to T\mathbb{R}^n\) given by \[ (p, V) \mapsto [p + tV]. \]
The projection \(\pi : TM \to M\) is the map \[ \pi([c]) = c(0). \]
Note that \(\pi\) is well defined: if \([c] = [d]\) so that \(c \sim d\), then \(c(0) = d(0)\) hence \(\pi([c]) = \pi([d])\). The projection takes a tangent vector as input and returns the basepoint.
The tangent bundle is a topological manifold of dimension \(2n\).
We will first define charts and then use them to define the topology. For any chart \(\varphi : U \subseteq M \to \mathbb{R}^n\) for \(M\), define the chart \[ \Phi : \pi^{\ast} (U) \subseteq TM \to \varphi(U) \times \mathbb{R}^n \subseteq \mathbb{R}^n \times \mathbb{R}^n \] for \(TM\) by \[ \Phi(V) = \big(\varphi \circ \pi(V), (\varphi \circ c)'(0)\big) \] where \(V = [c]\). Since \(\pi(V) = c(0)\) we may write this in the appealing way, \[ \Phi(V) = \big(\varphi \circ c(0), (\varphi \circ c)'(0)\big). \] That is, \(\Phi(V)\) identifies \(V\) with \((x, X)\) as a point and velocity vector on the curve \(\varphi \circ c\) which is how we think of tangent vectors in \(\mathbb{R}^n\).
Since \(V \in \pi^{\ast} (U)\), we have \(p = \pi(V) \in U\). Thus \(x = \varphi \circ \pi(V) = \varphi(p) \in \varphi(U)\) which is an open set in \(\mathbb{R}^n\). Moreover, \((\varphi \circ c)'(0)\) is well defined independently of the choice of \(c \in V\) (i.e. the choice of \(c\) such that \(V = [c]\) since by definition if \(c \sim d\), then \((\varphi \circ c)'(0) = (\varphi \circ d)'(0)\).
The set \(\varphi(U)\) is open by definition of charts on \(M\), hence the set \(\varphi(U) \times \mathbb{R}^n\) is an open subset of \(\mathbb{R}^n \times \mathbb{R}^n \simeq \mathbb{R}^{2n}\).
Covering \(M\) by charts \(\varphi_{\alpha} : U_{\alpha} \subseteq M \to V_{\alpha} \subseteq \mathbb{R}^n\), we obtain a cover of \(TM\) by \(\widetilde{U_{\alpha}} := \pi^{\ast} (U_{\alpha})\). Then we define a set \(\widetilde{U} \subseteq TM\) to be open if and only if for every \(\alpha\), \(\Phi_{\alpha}(\widetilde{U} \cap \widetilde{U}_{\alpha}) \subseteq \mathbb{R}^{2n}\) is open.
This definition defines a topology and (basically by definition), for every \(\alpha\), \(\Phi_{\alpha}\) is a homeomorphism. Hence \(TM\) is a topological manifold.
The requirements that \(TM\) be a topological manifold with charts \(\Phi_{\alpha}\) uniquely determines the topology on \(TM\).
In the definition of topological manifold, the domains \(\widetilde{U}_{\alpha}\) of the charts must be open. Next, each \(\Phi_{\alpha} : \widetilde{U}_{\alpha} \to U_{\alpha} \times \mathbb{R}^n\) must be a homeomorphism. We already know that \(\Phi_{\alpha}\) is a bijection. Then \(\Phi_{\alpha}\) is a homeomorphism precisely when it identifies open subsets of \(\widetilde{U}_{\alpha}\) with open subsets of \(U_{\alpha} \times \mathbb{R}^n\). Thus a subset \(V \subseteq \widetilde{U}_{\alpha}\) can be open if and only if \(\Phi_{\alpha}(V) \subseteq U_{\alpha} \times \mathbb{R}^n\) is open.
Since \(TM = \bigcup_{\alpha} \widetilde{U}_{\alpha}\), the sets \(\{\widetilde{U}_{\alpha}\}\) must form an open cover of \(TM\). For any open cover, and so in particular for our given open cover, a set \(\widetilde{U} \subseteq TM\) will be open if and only if for every \(\alpha\), \(\widetilde{U} \cap \widetilde{U}_{\alpha}\) is open. That is if and only if for every \(\alpha\), \(\Phi_{\alpha}(\widetilde{U} \cap \widetilde{U}_{\alpha})\) is open.
This precisely the definition we gave and the discussion here shows that it is the only possible definition!
Show that the open sets defined in the theorem form a topology. If you already know some topology, you just need to verify that the collection of all the sets \(\Phi_{\alpha}^{\ast}(W)\) is a basis for a topology. Otherwise, you need to show that \(\emptyset, TM\) are open according to the definition and that the collection of open sets is closed under arbitrary unions and finite intersections.
Prove that for every \(\alpha\), \(\Phi_{\alpha}\) is a homeomorphism. Hint : you need to show that \(\Phi_{\alpha}\) is continuous and \(\Phi_{\alpha}^{-1}\) is continuous. Since \(\Phi_{\alpha}\) is a bijection, this amounts to showing that \(\widetilde{U} \subseteq \widetilde{U}_{\alpha}\) is open if and only if \(\Phi_{\alpha}(\widetilde{U}_{\alpha})\) is open. Verify this latter statement and then verify that this statement is indeed true for our definition of open subsets of \(TM\).
The tangent bundle is a \(C^{\infty}\) manifold.
We just need to show that the transition maps \(T_{\alpha\beta} = \Phi_{\alpha} \circ \Phi_{\beta}^{-1}\) are \(C^{\infty}\).
If \(\widetilde{U}_{\alpha\beta} = \widetilde{U}_{\alpha} \cap \widetilde{U}_{\beta} = \emptyset\) there is nothing to prove.
Now for \((x, W) \in \varphi_{\beta}(U_{\alpha\beta}) \times \mathbb{R}^n \subseteq \mathbb{R}^{2n}\), we have \[ \Phi_{\beta}^{-1} (x, W) = [\varphi^{-1}_{\beta} (x + tW)] \in T_{\varphi(x)} M. \] Here \(\Phi_{\beta}^{-1} (x, W) = [c]\) with \(c(t) = \varphi_{\beta}^{-1}(x + tW)\) so that \[ \pi (\Phi_{\beta}^{-1} (x, W)) = c(0) = \varphi_{\beta}^{-1}(x). \] Applying \(\Phi_{\alpha}\) we get
\begin{align*} T_{\alpha\beta} (x, W) &= \Phi_{\alpha} \circ \Phi_{\beta}^{-1} (x, W) \\ &= \Phi_{\alpha}([c]) \\ &= \big(\varphi_{\alpha} \circ c(0), (\varphi_{\alpha} \circ c)'(0)\big) \\ &= \big(\varphi_{\alpha} \circ \varphi_{\beta}^{-1}(x), (\varphi_{\alpha} \circ \varphi_{\beta}^{-1} (x + tW))'(0)\big) \\ &= \big(\tau_{\alpha\beta}(x), d(\tau_{\alpha\beta})_x (W)\big) \end{align*}where \(\tau_{\alpha\beta} = \varphi_{\alpha} \circ \varphi_{\beta}^{-1}\) is the transition function for \(M\). Since \(\tau_{\alpha\beta}\) is \(C^{\infty}\), so to is \(x \mapsto d(\tau_{\alpha\beta})_x\). Since \(d(\tau_{\alpha\beta})_x(W)\) is linear in \(W\), it is \(C^{\infty}\) in \(W\) as well. Thus \(T_{\alpha\beta}\) is \(C^{\infty}\).