# 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 𝑀, then the tangent bundle 𝑇𝑀 is a smooth 2𝑛-manifold, how can I prove this?

• I assume you're familiar with how charts are chosen in $T(M)$, but you want to know why it is actually Hausdorff? This is what I got from the title, but it's not in the body of the question. – D_S Jun 24 '19 at 15:51

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.

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.