[proofplan] We prove the identity locally in a frame, where a connection is represented by a matrix-valued $1$-form $A$ and its curvature by $dA+A\wedge A$. The pulled-back connection has local matrix $f^*A$, so its curvature matrix is $d(f^*A)+f^*A\wedge f^*A=f^*(dA+A\wedge A)$. Since polynomial evaluation in Chern-Weil theory is built from wedge products, matrix multiplication, and scalar contractions, it commutes with pullback. [Closedness of Chern-Weil forms](/theorems/9757) then lets the equality pass to de Rham cohomology. [/proofplan]

[step:Write the connection and its pullback in a local frame] Let $U\subset M$ be an [open set](/page/Open%20Set) over which $E$ is trivial, and let \begin{align*} s=(s_1,\dots,s_r) \end{align*} be a smooth local frame for $E|_U$. Let $V:=f^{-1}(U)\subset N$, and let \begin{align*} f^*s=(f^*s_1,\dots,f^*s_r) \end{align*} denote the induced smooth local frame for $(f^*E)|_V$. Define the connection matrix of $\nabla$ in the frame $s$ to be the matrix-valued $1$-form \begin{align*} A=(A_{ij})\in \Omega^1(U;\mathfrak{gl}(r,\mathbb C)) \end{align*} specified by \begin{align*} \nabla s_j=\sum_{i=1}^r A_{ij}\otimes s_i. \end{align*} By the definition of the pulled-back connection, the connection matrix of $f^*\nabla$ in the frame $f^*s$ is \begin{align*} f^*A=(f^*A_{ij})\in \Omega^1(V;\mathfrak{gl}(r,\mathbb C)). \end{align*} Here we use the natural bundle isomorphism \begin{align*} \operatorname{End}(f^*E)\cong f^*\operatorname{End}(E), \end{align*} which sends an endomorphism of $E_{f(y)}$ to the corresponding endomorphism of $(f^*E)_y=E_{f(y)}$ for each $y\in N$. [/step]

[step:Compute that the curvature matrix pulls back]In the local frame $s$, the curvature of $\nabla$ is represented by the matrix-valued $2$-form \begin{align*} F_A=dA+A\wedge A\in \Omega^2(U;\mathfrak{gl}(r,\mathbb C)), \end{align*} where matrix multiplication is combined with the wedge product of forms. In the pulled-back frame $f^*s$, the curvature of $f^*\nabla$ is therefore represented by \begin{align*} F_{f^*A}=d(f^*A)+f^*A\wedge f^*A. \end{align*} Pullback of differential forms commutes with exterior differentiation and with wedge products, so \begin{align*} d(f^*A)=f^*(dA) \end{align*} and \begin{align*} f^*A\wedge f^*A=f^*(A\wedge A). \end{align*} Hence \begin{align*} F_{f^*A}=f^*(dA)+f^*(A\wedge A)=f^*(dA+A\wedge A)=f^*F_A. \end{align*} Thus, under $\operatorname{End}(f^*E)\cong f^*\operatorname{End}(E)$, \begin{align*} F_{f^*\nabla}=f^*F_\nabla. \end{align*}[/step]

[guided]The purpose of passing to a frame is to reduce the statement about bundle-valued forms to an explicit identity for matrices of ordinary differential forms. In the frame $s=(s_1,\dots,s_r)$ over $U$, the connection is encoded by a matrix-valued $1$-form $A=(A_{ij})$ through \begin{align*} \nabla s_j=\sum_{i=1}^r A_{ij}\otimes s_i. \end{align*} The curvature matrix in this same frame is \begin{align*} F_A=dA+A\wedge A. \end{align*} This formula uses the convention that $(A\wedge A)_{ij}=\sum_{\ell=1}^r A_{i\ell}\wedge A_{\ell j}$. Now restrict to $V=f^{-1}(U)$ and use the pulled-back frame $f^*s$. The pulled-back connection is defined so that its local connection matrix is $f^*A$. Therefore its curvature matrix is \begin{align*} F_{f^*A}=d(f^*A)+f^*A\wedge f^*A. \end{align*} We now use two standard naturality identities for differential forms: exterior differentiation commutes with pullback, and [pullback preserves wedge products](/theorems/3914). Applied entrywise to matrix-valued forms, these give \begin{align*} d(f^*A)=f^*(dA) \end{align*} and \begin{align*} f^*A\wedge f^*A=f^*(A\wedge A). \end{align*} Substituting these two identities into the curvature formula gives \begin{align*} F_{f^*A}=f^*(dA)+f^*(A\wedge A)=f^*(dA+A\wedge A)=f^*F_A. \end{align*} This is precisely the local expression of \begin{align*} F_{f^*\nabla}=f^*F_\nabla \end{align*} after identifying $\operatorname{End}(f^*E)$ with $f^*\operatorname{End}(E)$ fiberwise by the equality $(f^*E)_y=E_{f(y)}$.[/guided]

[step:Evaluate the invariant polynomial after pulling back the curvature] Write the polynomial $P$ as a finite sum of homogeneous components $P=\sum_{k=0}^m P_k$, where each $P_k$ is represented by its symmetric $k$-linear polarization $P_k:\mathfrak{gl}(r,\mathbb C)^k\to \mathbb C$. For an $\operatorname{End}(E)$-valued $2$-form $B$, the Chern-Weil evaluation is $P(B)=\sum_{k=0}^m P_k(B,\dots,B)$, where the expression is formed by wedging the form parts and contracting the matrix indices using $P_k$. Using $F_{f^*\nabla}=f^*F_\nabla$ and the fact that pullback commutes with wedge products and scalar contractions, for each $k$ we have $P_k(F_{f^*\nabla},\dots,F_{f^*\nabla})=P_k(f^*F_\nabla,\dots,f^*F_\nabla)=f^*P_k(F_\nabla,\dots,F_\nabla)$. Summing over $k$ gives $P(F_{f^*\nabla})=f^*P(F_\nabla)$ on $V=f^{-1}(U)$. Since the open sets $V$ obtained this way cover $N$, the equality holds globally on $N$. [/step]

[step:Pass the form identity to de Rham cohomology] By the closedness of Chern-Weil forms, as in [citetheorem:9757], both $P(F_\nabla)$ and $P(F_{f^*\nabla})$ are closed complex-valued differential forms. Therefore they define de Rham cohomology classes $[P(F_\nabla)]\in H^{\mathrm{even}}_{\mathrm{dR}}(M;\mathbb C)$ and $[P(F_{f^*\nabla})]\in H^{\mathrm{even}}_{\mathrm{dR}}(N;\mathbb C)$. The pullback map on de Rham cohomology is defined on a closed form $\alpha$ by $f^*[\alpha]=[f^*\alpha]$. Applying this definition to $\alpha=P(F_\nabla)$ and using the already proved equality of forms gives $[P(F_{f^*\nabla})]=[f^*P(F_\nabla)]=f^*[P(F_\nabla)]$. This proves the stated naturality identity both on forms and on de Rham cohomology classes. [/step]