[proofplan] Let $g = f^{-1}$ be the inverse map. Since $f$ is a Euclidean diffeomorphism, both $f$ and $g$ are differentiable, so the Euclidean chain rule applies to the identities $g \circ f = \operatorname{id}_U$ and $f \circ g = \operatorname{id}_V$. Differentiating these identities at the corresponding points shows that $Jg_{f(x)}$ is both a left and a right inverse of $Jf_x$. Hence $Jf_x$ is invertible. [/proofplan] [step:Introduce the inverse map and the two identity compositions] Fix $x \in U$, and define $y := f(x) \in V$. Since $f: U \to V$ is a Euclidean diffeomorphism, it is bijective and its inverse map \begin{align*} g: V \to U, \quad g(y') := f^{-1}(y') \end{align*} is smooth, hence differentiable, on $V$. The defining inverse identities are \begin{align*} g \circ f = \operatorname{id}_U \end{align*} and \begin{align*} f \circ g = \operatorname{id}_V. \end{align*} [/step] [step:Differentiate the identity $g \circ f = \operatorname{id}_U$ at $x$] The maps $f$ and $g$ are differentiable at $x$ and $f(x)=y$, respectively. By the Euclidean chain rule applied to $g \circ f: U \to U$ at $x$, \begin{align*} J(g \circ f)_x = Jg_{f(x)} Jf_x. \end{align*} Since $g \circ f = \operatorname{id}_U$, the [Jacobian matrix](/page/Jacobian%20Matrix) of $g \circ f$ at $x$ is the Jacobian matrix of the identity map on $U$, namely $I_n$. Therefore \begin{align*} Jg_y Jf_x = I_n. \end{align*} [guided] We first use the inverse identity that starts at $U$. Because $f$ is a Euclidean diffeomorphism, $f: U \to V$ is differentiable at $x$, and its inverse $g: V \to U$ is differentiable at the point $f(x)=y$. These are exactly the hypotheses needed for the Euclidean chain rule at $x$. Applying the chain rule to the composition \begin{align*} g \circ f: U \to U \end{align*} gives the matrix identity \begin{align*} J(g \circ f)_x = Jg_{f(x)} Jf_x. \end{align*} The order of multiplication matters: the differential of $f$ acts first, sending tangent vectors in $\mathbb{R}^n$ at $x$ to tangent vectors in $\mathbb{R}^n$ at $f(x)$, and then the differential of $g$ acts. Since $g=f^{-1}$, the composition $g \circ f$ is the identity map on $U$. The Jacobian matrix of the identity map $\operatorname{id}_U: U \to U$ at every point is the identity matrix $I_n$. Thus \begin{align*} Jg_y Jf_x = I_n. \end{align*} This says that $Jg_y$ is a left inverse of $Jf_x$. [/guided] [/step] [step:Differentiate the identity $f \circ g = \operatorname{id}_V$ at $y$] The maps $g$ and $f$ are differentiable at $y$ and $g(y)=x$, respectively. Applying the Euclidean chain rule to $f \circ g: V \to V$ at $y$ gives \begin{align*} J(f \circ g)_y = Jf_{g(y)} Jg_y. \end{align*} Since $f \circ g = \operatorname{id}_V$ and $g(y)=x$, we obtain \begin{align*} Jf_x Jg_y = I_n. \end{align*} [/step] [step:Conclude that the Jacobian matrix is invertible] Define the matrices \begin{align*} A := Jf_x \in \mathbb{R}^{n \times n} \end{align*} and \begin{align*} B := Jg_y \in \mathbb{R}^{n \times n}. \end{align*} The preceding two steps give \begin{align*} BA = I_n \end{align*} and \begin{align*} AB = I_n. \end{align*} Thus $B$ is a two-sided inverse of $A$. Therefore $A = Jf_x$ is invertible, with inverse \begin{align*} (Jf_x)^{-1} = Jg_y = J(f^{-1})_{f(x)}. \end{align*} Since $x \in U$ was arbitrary, $Jf_x$ is invertible for every $x \in U$. [/step]