[proofplan] We construct the correspondence in both directions. A section of $\operatorname{Hom}(E,F)$ evaluates on each vector in the corresponding fibre and therefore defines a fibrewise [linear map](/page/Linear%20Map) $E\to F$; local trivializations show this map is smooth. Conversely, a smooth bundle map covering $\operatorname{id}_M$ restricts on each fibre to a linear map, hence defines a point of the Hom fibre; again local trivializations show that these fibre maps vary smoothly. The two constructions are inverse because both are defined fibre by fibre by evaluation. [/proofplan]

[step:Send a Hom-bundle section to a smooth bundle map by fibrewise evaluation]Let $A\in \Gamma(M,\operatorname{Hom}(E,F))$ be a smooth section. Define the map $\Phi_A:E\to F$ by $\Phi_A(v)=A(\pi_E(v))(v)$ for every $v\in E$. For $v\in E_p$, where $p=\pi_E(v)$, we have $A(p)\in \operatorname{Hom}(E_p,F_p)$, so $A(p)(v)\in F_p$. Hence $\pi_F(\Phi_A(v))=p=\pi_E(v)$, and therefore $\Phi_A$ covers $\operatorname{id}_M$. For each $p\in M$, the fibre restriction $(\Phi_A)|_{E_p}:E_p\to F_p$ is the map $v\mapsto A(p)(v)$ and is linear because $A(p)$ is a linear map. It remains to verify smoothness. Let $p_0\in M$. Choose an open neighbourhood $U\subset M$ of $p_0$ over which both bundles trivialize, with smooth local trivializations $\tau_E:\pi_E^{-1}(U)\to U\times \mathbb{R}^r$ and $\tau_F:\pi_F^{-1}(U)\to U\times \mathbb{R}^s$, where $r=\operatorname{rank}E$ and $s=\operatorname{rank}F$. These trivializations induce a local trivialization \begin{align*} \tau_{\operatorname{Hom}}:\operatorname{Hom}(E,F)|_U&\to U\times \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s) \end{align*} by sending $B\in \operatorname{Hom}(E_p,F_p)$ to \begin{align*} (p,\tau_{F,p}\circ B\circ \tau_{E,p}^{-1}), \end{align*} where $\tau_{E,p}:E_p\to \mathbb{R}^r$ and $\tau_{F,p}:F_p\to \mathbb{R}^s$ are the fibre maps induced by $\tau_E$ and $\tau_F$. Since $A$ is smooth, its local representative is a smooth map \begin{align*} a:U&\to \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s) \end{align*} defined by \begin{align*} \tau_{\operatorname{Hom}}(A(p))=(p,a(p)). \end{align*} In the same trivializations, $\Phi_A$ is represented by the map $U\times \mathbb{R}^r\to U\times \mathbb{R}^s$ given by $(p,x)\mapsto (p,a(p)x)$. The map $(p,x)\mapsto a(p)x$ is smooth because matrix multiplication is a smooth finite-dimensional bilinear map and $a$ is smooth. Since $p_0$ was arbitrary, $\Phi_A$ is smooth on $E$.[/step]

[guided]We start with a smooth section $A$ of the Hom bundle. This means that for every point $p\in M$, the value $A(p)$ is a linear map $A(p):E_p\to F_p$. Thus the only possible bundle map associated to $A$ is obtained by evaluation on the fibre. Define $\Phi_A:E\to F$ by $\Phi_A(v)=A(\pi_E(v))(v)$ for every $v\in E$. This formula is well-defined because if $v\in E_p$, then $\pi_E(v)=p$, and $A(p)$ has domain $E_p$. Therefore $A(p)(v)\in F_p$. Now we verify that $\Phi_A$ is a vector bundle map covering $\operatorname{id}_M$. For $v\in E_p$, the image $\Phi_A(v)=A(p)(v)$ lies in $F_p$, so \begin{align*} \pi_F(\Phi_A(v))=p=\pi_E(v). \end{align*} Hence $\pi_F\circ \Phi_A=\pi_E$, which is exactly the statement that $\Phi_A$ covers $\operatorname{id}_M$. For fixed $p\in M$, the restriction $(\Phi_A)|_{E_p}:E_p\to F_p$ is the map $v\mapsto A(p)(v)$ and is linear because $A(p)$ is an element of $\operatorname{Hom}(E_p,F_p)$. The remaining point is smoothness. Smoothness of a bundle map is checked in local trivializations. Fix $p_0\in M$, and choose an open neighbourhood $U\subset M$ of $p_0$ over which both vector bundles are trivial. Let $\tau_E:\pi_E^{-1}(U)\to U\times \mathbb{R}^r$ and $\tau_F:\pi_F^{-1}(U)\to U\times \mathbb{R}^s$ be smooth local trivializations, where $r=\operatorname{rank}E$ and $s=\operatorname{rank}F$. These induce a trivialization of the Hom bundle by representing a fibre map $B:E_p\to F_p$ as the linear map between the model fibres. Explicitly, $\tau_{\operatorname{Hom}}:\operatorname{Hom}(E,F)|_U\to U\times \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$ sends $B$ to $(p,\tau_{F,p}\circ B\circ \tau_{E,p}^{-1})$. Here $\tau_{E,p}:E_p\to \mathbb{R}^r$ and $\tau_{F,p}:F_p\to \mathbb{R}^s$ are the fibrewise linear isomorphisms obtained from $\tau_E$ and $\tau_F$. Because $A$ is a smooth section, its local representative in this Hom-bundle trivialization is a smooth map \begin{align*} a:U&\to \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s) \end{align*} satisfying \begin{align*} \tau_{\operatorname{Hom}}(A(p))=(p,a(p)). \end{align*} Now take a vector $v\in E_p$ and write $\tau_E(v)=(p,x)$ with $x\in \mathbb{R}^r$. Since $a(p)$ represents the linear map $A(p)$ in the chosen trivializations, the image $\Phi_A(v)=A(p)(v)$ is represented in $F$ by \begin{align*} (p,a(p)x). \end{align*} Thus the local coordinate expression of $\Phi_A$ is the map $U\times \mathbb{R}^r\to U\times \mathbb{R}^s$ given by $(p,x)\mapsto (p,a(p)x)$. This map is smooth because $a:U\to \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)$ is smooth and the evaluation map $\operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s)\times \mathbb{R}^r\to \mathbb{R}^s$ given by $(T,x)\mapsto T(x)$ is a smooth finite-dimensional bilinear map. Since every point of $M$ has such a neighbourhood $U$, the map $\Phi_A:E\to F$ is smooth globally.[/guided]

[step:Send a bundle map covering the identity to a Hom-bundle section by fibre restriction] Let $\Phi\in \operatorname{Hom}_M(E,F)$ be a smooth vector bundle map covering $\operatorname{id}_M$. For each $p\in M$, define $A_\Phi(p):E_p\to F_p$ by $A_\Phi(p)(v)=\Phi(v)$ for every $v\in E_p$. Because $\Phi$ covers $\operatorname{id}_M$, if $v\in E_p$, then $\Phi(v)\in F_p$. Because $\Phi$ is a vector bundle map, the restriction $A_\Phi(p)=\Phi|_{E_p}$ is linear. Hence $A_\Phi(p)\in \operatorname{Hom}(E_p,F_p)$, so $A_\Phi$ is a set-theoretic section of $\operatorname{Hom}(E,F)\to M$. We verify that $A_\Phi$ is smooth. Let $p_0\in M$, and choose local trivializations $\tau_E:\pi_E^{-1}(U)\to U\times \mathbb{R}^r$ and $\tau_F:\pi_F^{-1}(U)\to U\times \mathbb{R}^s$ over an open neighbourhood $U\subset M$ of $p_0$. Since $\Phi$ is smooth, its local representative has the form $\phi:U\times \mathbb{R}^r\to U\times \mathbb{R}^s$ with $\phi(p,x)=(p,b(p,x))$ for a smooth map $b:U\times \mathbb{R}^r\to \mathbb{R}^s$. For each $p\in U$, the map $x\mapsto b(p,x)$ is linear because $\Phi|_{E_p}$ is linear. Let $(e_1,\dots,e_r)$ denote the standard basis of $\mathbb{R}^r$. Define \begin{align*} a:U&\to \operatorname{Hom}(\mathbb{R}^r,\mathbb{R}^s) \end{align*} by the requirement \begin{align*} a(p)e_i=b(p,e_i) \end{align*} for every $i\in\{1,\dots,r\}$. Since each component map $p\mapsto b(p,e_i)$ is smooth, the matrix entries of $a(p)$ are smooth functions of $p$, and therefore $a$ is smooth. In the induced Hom-bundle trivialization, \begin{align*} \tau_{\operatorname{Hom}}(A_\Phi(p))=(p,a(p)). \end{align*} Thus $A_\Phi$ is a smooth section of $\operatorname{Hom}(E,F)$. [/step]

[step:Check that the two fibrewise constructions are inverse to each other] Let $A\in \Gamma(M,\operatorname{Hom}(E,F))$. For every $p\in M$ and every $v\in E_p$, the section associated to the bundle map $\Phi_A$ satisfies \begin{align*} A_{\Phi_A}(p)(v)=\Phi_A(v)=A(p)(v). \end{align*} Since two linear maps $E_p\to F_p$ are equal exactly when they agree on every $v\in E_p$, we have $A_{\Phi_A}(p)=A(p)$ for every $p\in M$. Hence $A_{\Phi_A}=A$. Conversely, let $\Phi\in \operatorname{Hom}_M(E,F)$. For every $v\in E$, put $p=\pi_E(v)$. Then \begin{align*} \Phi_{A_\Phi}(v)=A_\Phi(p)(v)=\Phi(v). \end{align*} Therefore $\Phi_{A_\Phi}=\Phi$. The assignments $A\mapsto \Phi_A$ and $\Phi\mapsto A_\Phi$ are inverse maps between $\Gamma(M,\operatorname{Hom}(E,F))$ and $\operatorname{Hom}_M(E,F)$. This proves the natural bijection. [/step]