I am reading Lee's ''Introduction to Smooth Manifolds'' and it states that the tangent bundle TM has a natural topology and smooth structure that makes it a 2n-dimensional manifold.
I get most of the proof, my only problem is to check the Hausdorff condition because it does not specify which topology they use, so I thought of one that works, but I do not know if it is the usual one.
Having the specified topology does not bother most of the proof since you have a projection: $\pi: TM \to M$ defined as $\pi(p,X)=p$
And just take open sets as $\pi^{-1}(U)$ where $U$ is open in $M$. This works well for most of the proof, but when you want to check the Hausdorff condition in two points in the same fiber $(p,X),(p,Y)$.
So to avoid this I can define the open sets as the collection of sets:
$(U,\{V_x\}_{x \in U}):=\{ (x,v) : x \in U, v \in V_x \}$ where $U$ is an open set of $M$ and $\{V_x\}_{x \in U}$ a collection of open sets $V_x \subset T_xM$ where many but not all can be the empty set.
Is this right? I checked and it does not change the smooth structure since the local charts $(U,\phi^{\sim})$ can be seen as $((U, \{T_xM\}_{x\in M}),\phi^{\sim})$ in my topology. Thanks in advance.
Best Answer
You know that each chart $\phi : U \to V \subset \mathbb{R}^n$ for $M$ induces a canonical bijection $\tilde{\phi} : TM \mid_U = p^{-1}(U) \to V \times \mathbb{R}^n$. We define $W \subset TM$ to be open if $\tilde{\phi}(W \cap p^{-1}(U))$ is open in $V \times \mathbb{R}^n$ for all $\phi$.
It is an easy exercise to show that this is in fact a Hausdorff topology.