# The tangent space is well-defined

I'm trying to show that the tangent space $$T_xX$$ of a manifold $$X$$ at $$x$$ is well-defined. Could you have a check on my proof?

First, I recall related definitions to remove ambiguity.

• A subset $$X \subseteq \mathbb{R}^{N}$$ is called a smooth $$n$$-dimensional manifold if $$\forall x \in X$$, $$\exists$$ a diffeomorphism $$\varphi: U \to V$$ such that $$V$$ is open in $$X$$, $$U$$ is open in $$\mathbb{R}^{n}$$, and $$x \in V$$. Then $$\varphi$$ is called a local parameterization of $$V$$. The inverse $$\varphi^{-1}$$ is called a local coordinate system, or chart, on $$V$$.

• Let $$X \subseteq \mathbb R^N$$ be a $$n$$-dimensional smooth manifold and $$x \in X$$. Let $$\varphi: U \to V$$ be a local parameterization around $$x \in V$$. The continuous linear map $$\mathrm d \varphi_{\varphi^{-1}(x)} : \mathbb R^n \to \mathbb R^N$$ is the Fréchet derivative of $$\varphi$$ at $$\varphi^{-1}(x) \in U$$. The tangent space of $$X$$ at $$x$$, denoted by $$T_xX$$, is defined as the image of $$\mathrm d \varphi_{\varphi^{-1}(x)}$$, i.e., $$T_xX := \operatorname{im} \left (\mathrm d \varphi_{\varphi^{-1}(x)} \right ).$$

Theorem: Let $$X \subseteq \mathbb R^N$$ be a $$n$$-dimensional smooth manifold and $$x \in X$$. Then the tangent space $$T_xX$$ is well-defined, i.e. it's independent from the choice of local parameterization.

Proof: Let $$\varphi_1: U_1 \to V_1$$ and $$\varphi_2: U_2 \to V_2$$ be two local parameterizations around $$x$$. Let $$y_1 := \varphi^{-1}_1(x)$$ and $$y_2 := \varphi^{-1}_2(x)$$. Let $$V := V_1 \cap V_2$$, $$U_1 := \varphi_1^{-1}(V)$$, and $$U_2 := \varphi_2^{-1}(V)$$. Then $$V$$ is open in $$X$$. Also, $$U_1$$ and $$U_2$$ are open both in $$U$$ and in $$\mathbb R^n$$. Also, $$y_1 \in U_1$$ and $$y_2 \in U_2$$.

Let $$\phi_1: U_1 \to V$$ and $$\phi_2: U_2 \to V$$ be the restrictions of $$\varphi_1$$ on $$U_1$$ and $$\varphi_2$$ on $$U_2$$ respectively. Then they are also local parameterizations around $$x$$. Let $$\psi := \phi_1^{-1} \circ \phi_2$$. Then $$\psi^{-1} \circ \psi = \operatorname{id}_{U_2}$$ and $$\psi \circ \psi^{-1} = \operatorname{id}_{U_1}$$. Notice that $$\psi$$ is a diffeomorphism from $$U_2$$ to $$U_1$$, so $$\mathrm d \psi (y_2)$$ and $$\mathrm d \psi^{-1} (y_1)$$ are bijective and thus automorphisms of $$\mathbb R^n$$. This in turn implies $$\operatorname{im} (\mathrm d \phi_2 (y_2) \circ \mathrm d \psi^{-1} (y_1)) = \operatorname{im} (\mathrm d \phi_2 (y_2)).$$

On the other hand, $$\phi_1 = \phi_2 \circ \psi^{-1}$$. By chain rule, $$\mathrm d \phi_1 (y_1) = \mathrm d \phi_2 (y_2) \circ \mathrm d \psi^{-1} (y_1).$$

The result then easily follows.

Your proof is correct. Just a minor suggestion: You use the fact that if $$\varphi : U \to V$$ is a local parameterization around $$x \in V$$ and $$\phi : U' \to V'$$ is a local local parameterization around $$x \in V'$$ which is a restriction of $$\varphi$$, then $$d\varphi_{\varphi^{-1}(x)} = d\phi_{\phi^{-1}(x)}$$. This is quite obvious but should perhaps be mentioned.
• Thank you so much. That point is important. It justifies the transfer from original parameterizations to new parameterizations whose images are the same. This in turn allows us to obtain a diffeomorphism from $U_1$ and $U_2$. All the good things arise from this diffeomorphism. Nov 22, 2021 at 18:31