Show that the tangent bundle $TM $ of a variety $M$ is Hausdorff. I read a Lemma Introduction to smooth manifolds" which says that given a smooth n -manifold , then the tangent bundle  is a smooth 2-manifold, how can I prove this?
 A: Let $v$ and $w$ be two distinct points in $T(M)$.  Say that $v \in T_x(M)$ and $w \in T_y(M)$ for some $x, y \in M$.  If $x = y$, then $v$ and $w$ can definitely be separated by an open set in $T(M)$, since locally $v$ and $w$ are given by the last $n$-coordinates in $\mathbb R^n \times \mathbb R^n$.
If $x \neq y$, choose disjoint charts $U$ and $V$ containing $x$ and $y$, respectively.  We can do this because $M$ is Hausdorff.  Then
$$\bigcup\limits_{p \in U} T_p(M) \cong U \times \mathbb R^n$$
$$\bigcup\limits_{p \in V} T_p(M) \cong V \times \mathbb R^n$$
are disjoint open sets of $T(M)$ containing $v$ and $w$, respectively.
A: We need to study the 2 possibles cases:
Considering  $\Pi$, the application, such that:
$\Pi: E \rightarrow M$
If:

$\Pi(x) \neq \Pi(y)$

Since $M$ is Hausdorff there are $U,V$ disjoint, such that: $ \Pi (x) \in U $, $ \Pi (y) \in V$
Then $x \in \Pi ^{- 1}(U)$ and $y \in \Pi^{- 1}(V)$,
Also, $U,V$ are disjoint
$ U \cap V \neq \emptyset $
$ \Pi ^{- 1} (U) \cap \Pi^{- 1} (V)  \neq \emptyset. $
If:

$\Pi(x) = \Pi(y) = p$.

By definition for each point $p$, there exists a $p \in U $ neighbourhood of the point that preserves a fiber differencing: 
$ \theta: \Pi^{- 1} (U) \rightarrow U \times R^{ n} $.
Then we have:
$\theta(x) = (p, x^{'})$
$ \theta(y) = (p, y^{'})$
Where $ x^{'}, y^{'} $ are vectors in $ R^{n}$
Since $\theta $ is a diffeormorphism, $ x^{'}\neq y^{'} $.
Also $ R^{n}$ is Hausdorff.
The neighbourhoods $ H, U \in R^{n}$ such that $ x^{'}\in H, y^{'} \in V $
Therefore,
$(p, x^{'}) \in U \times H$ 
and 
$(p, y^{'}) \in U \times V$.
Where $ x \in \theta^{- 1} (U \times H)$ and $ y \in \theta^{- 1} (U \times V) $ are disjoint environments.
We concluded that the tangent bundle $TM$ of a manifold $M $ is Hausdorff.
