Wrong proof of $TM$ is diffeomorphic to $M\times \mathbb{R^m}$ I want to see what's wrong in here:
Let $M$ be a smooth manifold with dimension $m$. I will show $TM$ is diffeomorphic to $M\times \mathbb{R^m}$.
proof) Define $F:TM\rightarrow M\times \mathbb{R^m}$ by $F(p,v)=(p,v^1,...,v^m)$ where $v=v^i\frac{\partial}{\partial x^i}\in T_pM$.
Let $(U,\phi)$ be a chart containing $p$. Then, $(\pi^{-1}(U),\widetilde{\phi})$ is a chart containing $(p,v)$ where $\pi:TM\rightarrow M$ given by $\pi(p,v)=p$ and $\widetilde{\phi}(p,v)=(\phi(p),v^1,...,v^m)$. And $(U\times \mathbb{R^m},\phi \times Id)$ is a chart containing $F(p,v)$.
Using above, $(\phi\times Id)\circ F\circ \widetilde{\phi}^{-1}:\widetilde{\phi} (\pi^{-1}(U))\rightarrow \phi(U)\times \mathbb{R^m}$ is an identity map (Note that $\widetilde{\phi} (\pi^{-1}(U))$ is $\phi(U)\times \mathbb{R^m}$ by calculation.). Thus $F$ is smooth.
$F^{-1}:M\times \mathbb{R^m}\rightarrow TM$ is given by $F^{-1}(p,v^1,...,v^m)=(p,v)$ where $v=v^i\frac{\partial}{\partial x^i}\in T_pM$. With above charts, we have $\widetilde{\phi}\circ F^{-1}\circ (\phi\times Id)^{-1}:\phi(U)\times \mathbb{R^m}\rightarrow \widetilde{\phi}(\pi^{-1}(U))$ is also identity map. Thus $F^{-1}$ is smooth. $\blacksquare$
But I know $TM$ may not diffeomorphic to $M\times \mathbb{R^m}$. What's wrong in my proof?
 A: First of all, the $F$ you have defined is not actually a map on the tangent bundle of the manifold.The point is,when you are explicitly using coordinates to define a map,what you need to check is whether such a thing is independent of the chart chosen.The reason is that, the map you are actually defining takes a point on the manifold and gives you an output, but the same point on the manifold may have different coordinate representations.That is the reason why you cannot just use a chart and work your way with respect to it, it simply might not give you a map on the manifold at all.In your case, for example, if I use a different local chart at that point, then a point on $TM$ would end up having two different coordinate representations, what will $F$ be then?
Edit :- I thought I need to mention this.Another very common mistake with using charts to define maps is, a 'proof' that all smooth $n$-manifolds are orientable.Indeed, one just takes the 'everywhere positive $n$-form' given by $\omega = dx^1\wedge dx^2\wedge ..\wedge dx^n$.Spot the error. 
:-)
