[proofplan] The proof is a direct unpacking of definitions. Smoothness of $f$ means that all component functions have continuous partial derivatives of every order on $U$; in particular, the first-order partial derivatives exist and are continuous. Those first-order data are precisely the condition that $f$ belongs to $C^1(U;\mathbb{R}^m)$. [/proofplan]

[step:Unpack smoothness component by component] For each index $i \in \{1,\dots,m\}$, define the component map \begin{align*} f_i: U \to \mathbb{R} \end{align*} by declaring that \begin{align*} f(x) = (f_1(x),\dots,f_m(x)) \end{align*} for every $x \in U$. Since $f: U \to \mathbb{R}^m$ is smooth, each component $f_i: U \to \mathbb{R}$ is smooth. Therefore, for every $i \in \{1,\dots,m\}$ and every coordinate index $j \in \{1,\dots,n\}$, the first [partial derivative](/page/Partial%20Derivative) \begin{align*} \partial_{x_j} f_i: U \to \mathbb{R} \end{align*} exists and is continuous. [/step]

[step:Recognize the first derivative data as the definition of $C^1$]The definition of $C^1(U;\mathbb{R}^m)$ for a map $f: U \to \mathbb{R}^m$ requires that all first-order partial derivatives of all component functions exist and are continuous on $U$. The preceding step verifies exactly this condition for every component index $i \in \{1,\dots,m\}$ and every coordinate index $j \in \{1,\dots,n\}$. Hence \begin{align*} f \in C^1(U;\mathbb{R}^m). \end{align*}[/step]

[guided]We now match the information obtained from smoothness with the definition of the target space. The space $C^1(U;\mathbb{R}^m)$ consists of maps $f: U \to \mathbb{R}^m$ whose component functions have continuous first-order partial derivatives on $U$. From smoothness, we already know more: for each component map $f_i: U \to \mathbb{R}$ and each coordinate direction $x_j$, the map \begin{align*} \partial_{x_j} f_i: U \to \mathbb{R} \end{align*} exists and is continuous. This is the entire first-order regularity requirement. Since the condition holds for every $i \in \{1,\dots,m\}$ and every $j \in \{1,\dots,n\}$, the map $f$ has continuous first-order component derivatives on $U$. Therefore, by the definition of $C^1(U;\mathbb{R}^m)$, \begin{align*} f \in C^1(U;\mathbb{R}^m). \end{align*}[/guided]