TITLE: Let $G$ be a Lie group. Show that there is a diffeomorphism $TG \cong G \times T_e G$.
QUESTION [28 upvotes]: Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.
Thanks so much!

REPLY [4 votes]: This question is rather old. But since I think Alex Youcis solution is not quite right (or I'm interpreting it wrongly), I will post an answer with explicit computations that solves the case of the cotangent bundle too (by replacing $T$ by $T^{*}$), which cannot be solved by the categorial way presented by Alex Zorn (because multiplication turns into comultiplication!). Furthermore, often people consult this site, so this may be useful to anyone having difficults in this problem.  
About my notation, I will use $T$ istead of $d$ for the differential and $T_p(f)$ instead of $d(f)_p$.
Let $m: G \times G \longrightarrow G$ be the multiplication, $\pi: TG \twoheadrightarrow G$ the tangent bundle, $i: G \longrightarrow G$ the inversion and $$\Phi: TG \longrightarrow TG$$ given by $$\Phi (x) = L_{i(\pi (x))}x$$, where $T l_g = L_g: TG \longrightarrow TG$ and $l_g = m\circ i_{g}: G \longrightarrow G$ and $i_g : G \longrightarrow \{ g \} \times G \hookrightarrow G \times G$ the inclusion.
Let $\phi_i : U_i \longrightarrow \mathbb{R}^n$ be charts of $G$ and $\varphi: T\mathbb{R}^n \longrightarrow \mathbb{R}^n \times \mathbb{R}^n$ a diffeomorphism. In this case, $\varphi \circ T\phi_i: TU_i \longrightarrow \mathbb{R}^n \times \mathbb{R}^n$ is a chart.
Locally, $\Phi$ is written as $$\varphi \circ T\phi_j \circ \Phi \circ T \phi_i^{-1} \circ \varphi^{-1} (x, v) = $$ $$= \varphi \circ T\phi_j(L_{i(\pi ( T \phi_i^{-1} \circ \varphi^{-1} (x, v)))} ( T \phi_i^{-1} \circ \varphi^{-1} (x, v))) =$$ $$= \varphi \circ T\phi_j(T(m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))}))  (T \phi_i^{-1} \circ \varphi^{-1} (x, v))) =$$ $$ =\varphi \circ T (\phi_j \circ \phi_i^{-1} \circ m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))}\circ  \phi_i^{-1}) \circ \varphi^{-1} (x, v)$$. The unique term that need to be checked if it's indeed smooth is $m \circ i_{ i(\pi (T \phi_i^{-1} \circ \varphi^{-1} (x, v) ))}$. However, for any smooth morphism $\psi: P \longrightarrow G$ (where $P$ is any smooth manifold),  $m \circ i_{\psi (-)}: P \times G \longrightarrow G$ is smooth since $i_{\psi ( P)}$ is an inclusion for each $p$. Note that the derivative is not taken with respect to $m (\psi (-), -)$! In other words, for each $p$, $\psi (p) = g$ and just after this $m \circ i_{g} : G \longrightarrow G$ is differentiated. This is because we want $T(m\circ i_g) = L_g$ and not $T (m( \psi \times 1_G))_{p, g} = T_{(\psi( p ) ,g) } m (T_{p}\psi \times 1_{T_{g}G}) = L_{\psi (p )} (1_{T_{g}G}) + R_{g} T_p \psi$.
Now it remains to note that the map $\Phi: TG \longrightarrow TG$ have image in $T_e G$ since $L_{i (\pi (x))} : T_{\pi (x)} G \longrightarrow T_e G$. The surejctivity and injectivity follow easily by algebraic computations and the fact that $L_g$ is an isomorphism of vector bundles and the fact that this representation ($g \mapsto L_g$) of $G$ is a homomorphism.