# If $f :X \to Y$ is a diffeomorphism, then $df_x$ is an isomorphism of vector spaces.

${\bf Proposition}$: If $f: X \to Y$ is a diffeomorphism, then $df_x$ is an isomorphism of tangent spaces.

${\bf Proof}$: Let $f: X \to Y$ be a diffeomorphism. Then $f$ is a smooth bijection and so is $f^{-1} : Y \to X$. Since $f$ is a bijection, $\dim X = \dim Y$. Therefore, $\dim T_x(X) = \dim T_y(Y)$ for all $x \in X, y \in Y$. Take any $h_1, h_2 \in T_x(X)$ for some $x \in X$ and take $\alpha \in \mathbb{R}$. If $f(x) = y$, we need to show that $df_x(h_1 + h_2) = df_x(h_1) + df_x(h_2)$ and that $df_x(\alpha h_1) = \alpha df_x(h_1)$ and that $df_x:T_x(X) \to T_{y}(Y)$ is a bijection.

$df_x$ is a linear map so it is trivially true, by the definition of linearity, that it preserves the vector space structure between $T_x(X)$ and $T_y(Y)$.

To prove that it is a bijection, I need to show that $df_x(T_x(X)) = T_y(Y)$ and that if $df_x(h_1) = df_x(h_2)$, then $h_1 = h_2$. How can I go about proving these surjective and injective properties?

Thanks.

• You only need to show it is injective - it's a map between vector spaces of the same dimension. To this end, it's enough to check that the image of the basis vectors of one tangent space form a linearly independent set. Sep 8, 2016 at 1:39
• Do you know what the chain rule says? That gives the easiest proof of this statement that I know of. Sep 8, 2016 at 2:05
• One place where you go wrong is the statement "Since $f$ is a bijection, $\text{dim} \, X = \text{dim} \, Y$." There exists a bijection between $\mathbb{R}$ and $\mathbb{R}^2$, but they have different dimensions. Sep 8, 2016 at 12:27

If $f$ is a diffeomorphism then is has inverse say $g$. Let $p \in X$ and $f(p) \in Y$ then first recall that;

$$(f \circ g)_{*,f(p)} =f_{*,p} \circ g_{*,f(p)}$$

By definition of $g$ is follows that;

$$f_{*,p} \circ g_{*,f(p)} = (f \circ g)_{*,f(p)} = (\iota_Y)_{*,f(p)}$$

$$\hspace{-.3in} g_{*,f(p)} \circ f_{*,p} = (g \circ f)_{*,p} = (\iota_X)_{*,p}$$

$$\\$$

Remark: Here $f_{*,p} = d_pf$. I've become accustomed to the other notation.