Tangent bundle : is a manifold I have studied what a differentiable manifold is, and what a tangent space at a given point is, and read the proof that its dimension is equal to the dimension of the manifold. Here a tangent vector is a derivation on the germs of differentiable functions at a point.
Now I am trying to see that the tangent bundle is indeed a differentiable manifold. I get that for an open set $U$ in Eucliden space, it takes the form $U \times \mathbb R^n$ with the vector $\sum a_i x_i$ giving the tangent vector $\sum a_i \partial /\partial x_i$. Now the book says that it is "obvious" that these glue up and give a differentiable manifold.
I may be stupid; but I will be grateful if somebody can hold my hand through this "obvious" statement.
 A: Take two coordinate charts $(U,\varphi)$ and $(V,\psi)$ ; you know that your transition function $\psi \circ \varphi^{-1} : \varphi(U \cap V) \to \psi(U \cap V)$ is smooth. Given a map $f : M \to N$ between manifolds and $p \in M$, I denote by $D_pf : T_pM \to T_{f(p)} N$ the corresponding map and $Df : TM \to TN$ (if we take all the maps $D_pf$ and gather them into one big map ; at this point we can assume nothing about $Df$). Furthermore, for any $p \in U \cap V$, you have maps 
$$
T_{\varphi(p)}\varphi(U \cap V) \overset{D_{\varphi(p)} \varphi^{-1}}{\longrightarrow} T_pM \overset{D_p \psi}{\longrightarrow} T_{\psi(p)}\psi(U \cap V)
$$
and you know that $D_p \psi \circ D_{\varphi(p)} \varphi^{-1} = D_{\varphi(p)} (\psi \circ \varphi^{-1})$ (assuming you defined the differential map via pullback of derivations, in which case the equation above is obvious ; otherwise it is the chain rule, so I guess you could prove it too with whatever definition you have). The assumption that $M$ is a manifold tells us that the maps
$$
D\varphi(TU) \cap D\varphi(TV) = \varphi(U \cap V) \times \mathbb R^m \overset{\alpha}{\longrightarrow} \psi(U \cap V) \times \mathbb R^m = D\psi(TU) \cap D\psi(TV)
$$
are smooth, since the map $\alpha = (\psi \circ \varphi^{-1}) \times (D(\psi \circ \varphi^{-1}))$ is given by $(\varphi(p), v) \mapsto (\psi(p), J_{\psi \circ \varphi^{-1}}(\varphi(p)))$ (where $J_{\psi \circ \varphi^{-1}}$ is the Jacobian matrix of $\psi \circ \varphi^{-1}$, thus is smooth!). 
We conclude that the collection $(TU, D\varphi)$ for $(U,\varphi)$ a coordinate chart of $M$ forms an atlas for $TM$, which means $TM$ is a manifold. Note that it is not hard to check in coordinates that the projection $TM \to M$ is smooth. 
Hope that helps (even though I'm a bit late, perhaps it will help others),
A: First, prove the following "topological glueing lemma":
Let $M$ be a set, and suppose that $M = \bigcup_\alpha U_\alpha$ for some subsets $U_\alpha$. Suppose moreover that each $U_\alpha$ is a topological space, that $U_\alpha \cap U_\beta$ is open in $U_\alpha$ for each $\alpha, \beta$, and that the two topologies on $U_\alpha \cap U_\beta$ induced by the inclusions $U_\alpha \cap U_\beta \subseteq U_\alpha$ and $U_\alpha \cap U_\beta \subseteq U_\beta$ coincide, for each $\alpha, \beta$. Then there is a unique topology on $M$ in which each $U_\alpha$ is open and such that the restriction of the topology of $M$ to each $U_\alpha$ is the topology of $U_\alpha$.
Then, prove the following "smooth glueing lemma":
Let $M$ be a set, and suppose that $M = \bigcup_\alpha U_\alpha$ for some subsets $U_\alpha$. Suppose moreover that each $U_\alpha$ is a smooth manifold, that $U_\alpha \cap U_\beta$ is open in $U_\alpha$ for each $\alpha, \beta$, and that the two smooth structures on $U_\alpha \cap U_\beta$ induced by the inclusions $U_\alpha \cap U_\beta \subseteq U_\alpha$ and $U_\alpha \cap U_\beta \subseteq U_\beta$ coincide, for each $\alpha, \beta$. Endow $M$ with the topology given by the topological glueing lemma. Then there is a unique smooth structure on $M$ whose restriction to each $U_\alpha$ is the smooth structure of $U_\alpha$.
