# The differential of any chart on a manifold $M$ is an isomorphism of $M$'s tangent space

My book asserts that a chart $(U,h)$ on an $n$-dimensional manifold $M$ induces an isomorphism of it's tangent space at a point $p$ and the tangent space of $\mathbb R^n$ at the image $h(p)$ via it's differential $T_ph:T_pU\to T_{h(p)}h[U]$.

How is it an isomorphism of vector spaces? Is it a diffeomorphism? Is the map on the tangent bundle $Th:TM\to T\mathbb R$ an isomorphism of vector spaces or a diffeomorphism of manifolds?

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Yes, it's because $h$ is a diffeomorphism, and the tangent space is functorial. – Alex Youcis Mar 30 '14 at 17:23
$T_p h$ is an isomorphism of vector space for each $p$ – Thomas Mar 30 '14 at 17:25
What do you mean that the tangent space is functorial, Mr. @Youcis. I thought we use the word functorial only when we talk about maps or functors. – Student Mar 30 '14 at 17:26
@Student The "taking of tangent space" operation is a functor, from smooth manifolds to vector spaces! But how do you define a tangent space? As a certain set of functions acting on germs? – Mike Miller Mar 30 '14 at 18:21
From pointed smooth manifolds. Sorry. – Mike Miller Mar 30 '14 at 18:28

If $f : L \to M$ is a smooth map, then for every $p \in L$ we have a corresponding linear map $df_p : T_pL \to T_pM$, where $df_p(X_p)\alpha = X_p(\alpha\circ f)$. Furthermore, we have the following properties:

• if $f : L \to M$, $g : M \to N$, then $d(g\circ f)_p = dg_{f(p)}\circ df_p$, and
• $d(\operatorname{id_M})_p = \operatorname{id}_{T_pM}$.

It follows that if $f$ is a diffeomorphism $M \to M$, then $df_p$ is an isomorphism $T_pM \to T_{f(p)}M$.

Now if $(U, \varphi)$ is a coordinate chart on $M$, then $\varphi$ is a diffeomorphism $U \to \varphi(U)$ where $\varphi(U)$ is an open subset of $\mathbb{R}^n$. Therefore $d\varphi_p : T_pU \to T_{\varphi(p)}\varphi(U)$ is an isomorphism so

$$T_pU \cong T_{\varphi(p)}\varphi(U) \cong T_{\varphi(p)}\mathbb{R}^n \cong \mathbb{R}^n.$$

In addition to the pointwise isomorphism of tangent spaces, we get a vector bundle isomorphism $d\varphi : TM|_U \to T\varphi(U)$. In particular, as $T\varphi(U) \cong \varphi(U)\times \mathbb{R}^n$, the tangent bundle to $M$, when restricted to a coordinate chart, is trivial. That is, the tangent bundle is trivialised by coordinate charts.

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