[proofplan] Work in the chosen local trivializations and write the coordinate representative of $\Phi$ as a map from $U\times\mathbb R^r$ to $V\times\mathbb R^s$. The condition that $\Phi$ lies over $f$ forces the base component to be $f(x)$, while fibrewise linearity says that the fibre component is, for each fixed $x$, a [linear map](/page/Linear%20Map) $\mathbb R^r\to\mathbb R^s$. The columns of this linear map are obtained by evaluating the fibre component on the standard basis vectors of $\mathbb R^r$, with the empty-basis convention when $r=0$, which proves smoothness of the matrix entries when $\Phi$ is smooth. Conversely, a smooth matrix-valued map gives a smooth coordinate formula because its fibre component is a finite sum of products of smooth coordinate functions, interpreted as the unique zero-dimensional expression in the rank-zero cases. [/proofplan]

[step:Pass to the coordinate representative of the bundle map] Let \begin{align*} \tau_E:E|_U\to U\times\mathbb R^r \end{align*} and \begin{align*} \tau_F:F|_V\to V\times\mathbb R^s \end{align*} denote the chosen local trivializations. Define the coordinate representative \begin{align*} \Psi:U\times\mathbb R^r\to V\times\mathbb R^s \end{align*} of $\Phi|_{E|_U}$ by \begin{align*} \Psi:=\tau_F\circ \Phi|_{E|_U}\circ \tau_E^{-1}. \end{align*} Since $\tau_E$ and $\tau_F$ are diffeomorphisms onto their local product charts, the Smooth Map coordinate criterion gives that $\Phi|_{E|_U}$ is smooth if and only if $\Psi$ is smooth. Let $B$ be the second component of $\Psi$, so $B$ is the map \begin{align*} B:U\times\mathbb R^r\to \mathbb R^s. \end{align*} Because $\Phi$ is a map over $f$, the first component of $\Psi$ is $f(x)$. Hence \begin{align*} \Psi(x,v)=(f(x),B(x,v)) \end{align*} for every $x\in U$ and $v\in\mathbb R^r$. [/step]

[step:Extract the smooth matrix entries from a smooth fibrewise linear map]Assume first that $\Phi|_{E|_U}$ is smooth. Then $\Psi$ is smooth, so its second component $B:U\times\mathbb R^r\to\mathbb R^s$ is smooth. For each $x\in U$, define $B_x:\mathbb R^r\to\mathbb R^s$ by $B_x(v)=B(x,v)$ for $v\in\mathbb R^r$. Fibrewise linearity of $\Phi$ implies that $B_x$ is linear for every $x\in U$. If $r>0$, let $e_1,\dots,e_r\in\mathbb R^r$ be the standard basis vectors; if $r=0$, this is the empty basis of the zero [vector space](/page/Vector%20Space) $\mathbb R^0$. For each $j\in\{1,\dots,r\}$, define $a_j:U\to\mathbb R^s$ by $a_j(x)=B(x,e_j)$ for $x\in U$. When $r=0$ there are no such maps $a_j$, and all subsequent column and sum statements are interpreted by the empty-column and empty-sum conventions. The map $a_j$ is smooth because it is the composition of the smooth map $x\mapsto (x,e_j)$ with the smooth map $B$. Define $A:U\to\mathbb R^{s\times r}$ by declaring that the $j$-th column of $A(x)$ is $a_j(x)$. Equivalently, if $a_j(x)=(A_{1j}(x),\dots,A_{sj}(x))$, then \begin{align*} A(x)=\bigl(A_{ij}(x)\bigr)_{1\le i\le s,\ 1\le j\le r}. \end{align*} Each coordinate function $A_{ij}:U\to\mathbb R$ is smooth, since it is a component of the smooth map $a_j$. Therefore $A$ is smooth. For $v=(v_1,\dots,v_r)\in\mathbb R^r$, linearity of $B_x$ gives \begin{align*} B(x,v)=B_x\left(\sum_{j=1}^r v_j e_j\right)=\sum_{j=1}^r v_j B_x(e_j)=\sum_{j=1}^r v_j a_j(x)=A(x)v. \end{align*} Thus the coordinate representative satisfies \begin{align*} \Psi(x,v)=(f(x),A(x)v) \end{align*} for every $x\in U$ and $v\in\mathbb R^r$, which is exactly the asserted formula for $\Phi$ in the chosen local product coordinates.[/step]

[guided]Assume that $\Phi|_{E|_U}$ is smooth. Since smoothness is a local coordinate property and $\tau_E,\tau_F$ are smooth trivializations with smooth inverses, the coordinate representative $\Psi:U\times\mathbb R^r\to V\times\mathbb R^s$ defined by \begin{align*} \Psi=\tau_F\circ \Phi|_{E|_U}\circ \tau_E^{-1} \end{align*} is smooth. Write the second component of $\Psi$ as a map \begin{align*} B:U\times\mathbb R^r\to\mathbb R^s. \end{align*} Because $\Phi$ lies over $f$, the base point of $\Phi(x,v)$ is $f(x)$. Therefore the first component of $\Psi(x,v)$ is forced to be $f(x)$, and we may write \begin{align*} \Psi(x,v)=(f(x),B(x,v)). \end{align*} Now fix $x\in U$. Define $B_x:\mathbb R^r\to\mathbb R^s$ by $B_x(v)=B(x,v)$ for $v\in\mathbb R^r$. The fibrewise linearity hypothesis says exactly that $B_x$ is linear. If $r>0$, a linear map $\mathbb R^r\to\mathbb R^s$ is determined by its values on the standard basis vectors $e_1,\dots,e_r\in\mathbb R^r$. If $r=0$, the standard basis is empty and there is a unique linear map from $\mathbb R^0$ to $\mathbb R^s$, represented by the unique $s\times 0$ matrix. Hence the natural way to recover the matrix of $B_x$ is to use those basis values as columns, with no columns in the case $r=0$. For each $j\in\{1,\dots,r\}$, define $a_j:U\to\mathbb R^s$ by $a_j(x)=B(x,e_j)$ for $x\in U$. When $r=0$, there are no indices $j$ and hence no maps $a_j$; the construction below produces the unique smooth map from $U$ to the space of $s\times 0$ matrices. For $r>0$, this map is smooth because $x\mapsto (x,e_j)$ is smooth from $U$ to $U\times\mathbb R^r$, and $B$ is smooth. Define $A:U\to\mathbb R^{s\times r}$ by taking $a_j(x)$ as the $j$-th column of $A(x)$. In coordinates, if \begin{align*} a_j(x)=(A_{1j}(x),\dots,A_{sj}(x)), \end{align*} then \begin{align*} A(x)=\bigl(A_{ij}(x)\bigr)_{1\le i\le s,\ 1\le j\le r}. \end{align*} Each entry $A_{ij}$ is smooth because it is a component function of $a_j$, so $A$ is a smooth map into $\mathbb R^{s\times r}$. Finally, for any vector $v=(v_1,\dots,v_r)\in\mathbb R^r$, the standard basis expansion gives \begin{align*} v=\sum_{j=1}^r v_j e_j. \end{align*} Using the linearity of $B_x$, we obtain \begin{align*} B(x,v)=B_x(v)=B_x\left(\sum_{j=1}^r v_j e_j\right)=\sum_{j=1}^r v_j B_x(e_j)=\sum_{j=1}^r v_j a_j(x)=A(x)v. \end{align*} Therefore the coordinate representative satisfies \begin{align*} \Psi(x,v)=(f(x),A(x)v). \end{align*} Equivalently, $\Phi$ has the asserted formula in the chosen product coordinates.[/guided]

[step:Prove smoothness from a smooth matrix formula] Conversely, suppose there is a smooth map $A:U\to\mathbb R^{s\times r}$ such that \begin{align*} \Phi(x,v)=(f(x),A(x)v) \end{align*} for every $x\in U$ and $v\in\mathbb R^r$. Write $A(x)=\bigl(A_{ij}(x)\bigr)_{1\le i\le s,\ 1\le j\le r}$ and $v=(v_1,\dots,v_r)$, with the understanding that this coordinate list is empty when $r=0$. The fibre component $C:U\times\mathbb R^r\to\mathbb R^s$ defined by $C(x,v)=A(x)v$ has coordinate functions \begin{align*} C_i(x,v)=\sum_{j=1}^r A_{ij}(x)v_j,\qquad i\in\{1,\dots,s\}, \end{align*} where the sum is the empty sum when $r=0$. If $s=0$, the target $\mathbb R^0$ has no coordinate functions, and $C$ is the unique map to the zero-dimensional vector space. Each $A_{ij}$ is smooth on $U$, each coordinate projection $(x,v)\mapsto v_j$ is smooth on $U\times\mathbb R^r$, and the standard closure properties in the definition of a Smooth Map imply that finite sums and products of smooth real-valued functions are smooth. Hence $C$ is smooth. Since $f:U\to V$ is smooth by hypothesis and $C$ is smooth, the coordinate characterization of a Smooth Map implies that the product map $\Psi:U\times\mathbb R^r\to V\times\mathbb R^s$ defined by $\Psi(x,v)=(f(x),C(x,v))$ is smooth. Thus the coordinate representative of $\Phi|_{E|_U}$ is smooth, and therefore $\Phi|_{E|_U}$ is smooth. [/step]

[step:Conclude the equivalence in the chosen trivializations] The previous two steps prove both implications. If $\Phi|_{E|_U}$ is smooth, then its coordinate representative has the form \begin{align*} (x,v)\mapsto (f(x),A(x)v) \end{align*} for a smooth map $A:U\to\mathbb R^{s\times r}$. Conversely, any smooth map $A:U\to\mathbb R^{s\times r}$ satisfying this coordinate formula makes the coordinate representative smooth, hence makes $\Phi|_{E|_U}$ smooth. This proves the stated local coordinate characterization of a vector bundle morphism. [/step]