[proofplan]
The pullback bundle has fibre $(f^*E)_x=\{x\}\times E_{f(x)}$, so each fibre is canonically identified with $E_{f(x)}$ by forgetting the first coordinate. Taking duals and exterior powers of this fibrewise identification gives the proposed maps. The only point requiring verification is that these fibrewise linear maps vary smoothly with $x$; this is checked in a local frame, where both maps are represented by identity matrices on the corresponding dual and exterior-power frames.
[/proofplan]
[step:Define the fibrewise dual map and verify that it is a linear isomorphism]
For each $x\in M$, define the projection-on-the-second-factor [linear map](/page/Linear%20Map) $\iota_x:(f^*E)_x\to E_{f(x)}$ by $\iota_x(x,e)=e$. This is a linear isomorphism of real vector spaces. Define the bundle map $\Phi:f^*(E^*)\to (f^*E)^*$ fibrewise as follows: for $x\in M$ and $\lambda\in E_{f(x)}^*$, set $\Phi(x,\lambda)=\lambda\circ \iota_x$. Equivalently,
\begin{align*}
\Phi(x,\lambda)(x,e)=\lambda(e)
\end{align*}
for every $e\in E_{f(x)}$.
For fixed $x\in M$, define $\Phi_x:(E_{f(x)})^*\to ((f^*E)_x)^*$ by $\Phi_x(\lambda)=\lambda\circ \iota_x$. This map is linear. Since $\iota_x$ is a linear isomorphism, precomposition with $\iota_x$ is a linear isomorphism of dual spaces. Its inverse sends a functional $\mu\in ((f^*E)_x)^*$ to the functional $E_{f(x)}\to \mathbb R$ defined by $e\mapsto \mu(x,e)$.
Thus $\Phi$ is a fibrewise linear bijection over $M$.
[/step]
[step:Define the fibrewise exterior-power map and verify that it is a linear isomorphism]
For each $x\in M$, the linear isomorphism $\iota_x:(f^*E)_x\to E_{f(x)}$ induces a linear isomorphism on exterior powers in the opposite displayed direction by sending vectors in $E_{f(x)}$ to their corresponding vectors in $(f^*E)_x$. Define the linear map $\Psi_{k,x}:\Lambda^k(E_{f(x)}) \to \Lambda^k((f^*E)_x)$ by declaring
\begin{align*}
\Psi_{k,x}(e_1\wedge \cdots \wedge e_k)=(x,e_1)\wedge \cdots \wedge (x,e_k)
\end{align*}
for $e_1,\dots,e_k\in E_{f(x)}$, and then extending linearly. This is well-defined because it is precisely the exterior power of the inverse linear isomorphism $\iota_x^{-1}:E_{f(x)}\to (f^*E)_x$ defined by $\iota_x^{-1}(e)=(x,e)$. Since $\Psi_{k,x}=\Lambda^k(\iota_x^{-1})$, its inverse is $\Lambda^k(\iota_x)$, which sends $(x,e_1)\wedge\cdots\wedge(x,e_k)$ to $e_1\wedge\cdots\wedge e_k$. Hence $\Psi_{k,x}$ is a linear isomorphism.
Define $\Psi_k:f^*(\Lambda^k E)\to \Lambda^k(f^*E)$ by $\Psi_k(x,\omega)=\Psi_{k,x}(\omega)$ for $x\in M$ and $\omega\in \Lambda^k(E_{f(x)})$. For $k=0$, both fibres are canonically $\mathbb R$, and $\Psi_{0,x}$ is the identity map on $\mathbb R$ by the convention $\Lambda^0 V=\mathbb R$ for every real [vector space](/page/Vector%20Space) $V$.
[/step]
[step:Check smoothness in local frames]
Let $U\subset N$ be an [open set](/page/Open%20Set) over which $E$ has a smooth local frame $s_1,\dots,s_r$, where each $s_i:U\to E$ is a smooth section and $s_1(y),\dots,s_r(y)$ is a basis of $E_y$ for each $y\in U$. Define $V=f^{-1}(U)\subset M$. For each $1\le i\le r$, define the pulled-back local frame section $\widetilde{s}_i:V\to f^*E$ by $\widetilde{s}_i(x)=(x,s_i(f(x)))$.
Since $f$ and $s_i$ are smooth, $\widetilde{s}_i$ is smooth, and $\widetilde{s}_1,\dots,\widetilde{s}_r$ is a frame for $f^*E$ over $V$.
Let $s_1^*,\dots,s_r^*$ denote the dual frame of $E^*$ over $U$, so
\begin{align*}
s_i^*(y)(s_j(y))=\delta_{ij}
\end{align*}
for all $y\in U$. The pullback frame of $f^*(E^*)$ over $V$ is given by the sections $x\mapsto (x,s_i^*(f(x)))$. The dual frame of $(f^*E)^*$ over $V$ is denoted by $\widetilde{s}_1^*,\dots,\widetilde{s}_r^*$ and is characterized by
\begin{align*}
\widetilde{s}_i^*(x)(\widetilde{s}_j(x))=\delta_{ij}.
\end{align*}
For every $x\in V$,
\begin{align*}
\Phi(x,s_i^*(f(x)))(\widetilde{s}_j(x))
=s_i^*(f(x))(s_j(f(x)))
=\delta_{ij}.
\end{align*}
Hence
\begin{align*}
\Phi(x,s_i^*(f(x)))=\widetilde{s}_i^*(x).
\end{align*}
Thus, in these local frames, $\Phi$ is represented by the identity matrix, so $\Phi$ is smooth and its fibrewise inverse is smooth.
For the exterior-power map, let $I=(i_1,\dots,i_k)$ range over strictly increasing $k$-tuples with $1\le i_1<\cdots<i_k\le r$. Define $s_I:U\to \Lambda^k E$ by $s_I(y)=s_{i_1}(y)\wedge\cdots\wedge s_{i_k}(y)$. These sections form a smooth local frame of $\Lambda^k E$. Define $\widetilde{s}_I:V\to \Lambda^k(f^*E)$ by $\widetilde{s}_I(x)=\widetilde{s}_{i_1}(x)\wedge\cdots\wedge \widetilde{s}_{i_k}(x)$. These sections form a smooth local frame of $\Lambda^k(f^*E)$. By the definition of $\Psi_k$,
\begin{align*}
\Psi_k(x,s_I(f(x)))=\widetilde{s}_I(x).
\end{align*}
Therefore $\Psi_k$ is represented by the identity matrix in the corresponding local exterior-power frames, so $\Psi_k$ is smooth and its fibrewise inverse is smooth.
[guided]
The smoothness issue is the only part not settled by fibrewise linear algebra. A map between vector bundles can be checked locally in frames: if its coordinate matrix in smooth local frames has smooth entries, then the bundle map is smooth. Here the coordinate matrices will be identity matrices.
Choose an open set $U\subset N$ on which $E$ admits a smooth local frame $s_1,\dots,s_r$, with each $s_i:U\to E$ smooth and with $s_1(y),\dots,s_r(y)$ a basis of $E_y$ for every $y\in U$. Set $V=f^{-1}(U)$. Since $f:M\to N$ is smooth, $V$ is open in $M$. Define $\widetilde{s}_i:V\to f^*E$ by $\widetilde{s}_i(x)=(x,s_i(f(x)))$.
These sections are smooth because they are obtained by composing the smooth map $f$ with the smooth frame sections $s_i$, and they form a frame because every fibre $(f^*E)_x$ is $\{x\}\times E_{f(x)}$.
Now consider the dual bundles. Let $s_1^*,\dots,s_r^*$ be the dual frame of $E^*$ over $U$, characterized by
\begin{align*}
s_i^*(y)(s_j(y))=\delta_{ij}
\end{align*}
for all $y\in U$. The corresponding pulled-back frame of $f^*(E^*)$ consists of the sections $x\mapsto (x,s_i^*(f(x)))$. The dual frame of $(f^*E)^*$ is denoted by $\widetilde{s}_1^*,\dots,\widetilde{s}_r^*$ and satisfies
\begin{align*}
\widetilde{s}_i^*(x)(\widetilde{s}_j(x))=\delta_{ij}.
\end{align*}
Using the definition of $\Phi$, for every $x\in V$ and every $1\le j\le r$,
\begin{align*}
\Phi(x,s_i^*(f(x)))(\widetilde{s}_j(x))=\Phi(x,s_i^*(f(x)))(x,s_j(f(x)))=s_i^*(f(x))(s_j(f(x)))=\delta_{ij}.
\end{align*}
Thus $\Phi(x,s_i^*(f(x)))$ is exactly $\widetilde{s}_i^*(x)$. In the chosen frames, $\Phi$ sends each basis vector to the corresponding basis vector, so its coordinate matrix is the identity matrix. The identity matrix has smooth constant entries, and its inverse is again the identity matrix. Hence $\Phi$ is a smooth vector bundle isomorphism.
The same local-frame computation proves smoothness for exterior powers. For each strictly increasing $k$-tuple $I=(i_1,\dots,i_k)$, define $s_I:U\to \Lambda^k E$ by $s_I(y)=s_{i_1}(y)\wedge\cdots\wedge s_{i_k}(y)$. The sections $s_I$ form the standard exterior-power frame of $\Lambda^k E$. The corresponding frame of $\Lambda^k(f^*E)$ is given by the sections $\widetilde{s}_I:V\to \Lambda^k(f^*E)$ defined by $\widetilde{s}_I(x)=\widetilde{s}_{i_1}(x)\wedge\cdots\wedge \widetilde{s}_{i_k}(x)$.
By the defining formula for $\Psi_k$,
\begin{align*}
\Psi_k(x,s_I(f(x)))=\Psi_k(x,s_{i_1}(f(x))\wedge\cdots\wedge s_{i_k}(f(x)))=(x,s_{i_1}(f(x)))\wedge\cdots\wedge (x,s_{i_k}(f(x)))=\widetilde{s}_{i_1}(x)\wedge\cdots\wedge \widetilde{s}_{i_k}(x)=\widetilde{s}_I(x).
\end{align*}
Again the local coordinate matrix is the identity matrix. Therefore $\Psi_k$ is smooth, and the same identity-coordinate argument applies to its inverse. For $k=0$, both exterior-power bundles are the product line bundle in these local descriptions, and the map is the identity on each fibre.
[/guided]
[/step]
[step:Conclude that the fibrewise identifications are canonical vector bundle isomorphisms]
The maps $\Phi$ and $\Psi_k$ cover the identity map on $M$, are linear isomorphisms on every fibre, and are smooth with smooth inverses by the local-frame computation. Hence they are smooth vector bundle isomorphisms over $M$.
The construction used only the defining fibre of the pullback bundle and the functorial operations of dualization and exterior power on vector spaces. Therefore the isomorphisms are canonical. This proves
\begin{align*}
f^*(E^*)\cong (f^*E)^*
\end{align*}
and
\begin{align*}
f^*(\Lambda^k E)\cong \Lambda^k(f^*E),
\end{align*}
with the stated fibrewise formulas.
[/step]
