Let $\Phi:A_M\longrightarrow A_N$ be a morphism of Lie algebroids covering $\phi:M\longrightarrow N$. Suppose $\Theta$ is a section of the pullback bundle $\phi^* A_N$.
How to compute $$\langle d_{A_N} \alpha_N, \Theta\wedge \Phi\circ \alpha\rangle$$ where $\alpha_N\in \Omega^1(A_N)=\mathsf{Hom}(\Gamma(A_N), C^\infty(N))=\Gamma(A_N^*)$ and $\alpha\in \Gamma(A_N)$ ?
Above $\langle -, -\rangle$ is the pairing $\langle -, -\rangle: \Omega^1(A_N)\times \Gamma(A_N)\longrightarrow C^\infty(N)$ defined pointwise by $$\langle \alpha_N, \alpha\rangle(p):=(\alpha_N)_p(\alpha(p)).$$
Recall, $d_{A_N}: \Omega^k(A_N)\longrightarrow \Omega^{k+1}(A_N)$ is defined by $$d_{A_N} \alpha_N(a_0, \ldots, a_{k})=\sum_{j=0}^k \mathcal{L}_{\sharp \alpha_j}(\alpha_N(a_0, \ldots, \widehat{\alpha_j}, \ldots, a_{k})-\sum_{i<j} \alpha_N([a_i, a_j],a_0, \ldots, \widehat{a_i}, \ldots, \widehat{a_j}, \ldots, a_k),$$ where $\sharp$ stands for the anchor of $A_N$. Again $$\Omega^k(A_N)=\mathsf{Hom}(\Lambda^k \Gamma(A_N), C^\infty(N))\simeq \mathsf{Alt}^k(\Gamma(A_N), C^\infty(N))\simeq \Gamma(\Lambda^k A_N^*).$$
Thanks.
