Let $A\longrightarrow M$ and $B\longrightarrow N$ be Lie algebroids with anchors $\#_A$ and $\#_B$, respectively.

A morphism of Lie algebroids is a morphism of vector bundles $\Phi:A\longrightarrow B$ covering let's say $\phi:M\longrightarrow N$ such that the induced map $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$ commutes with the exterior derivative. How can I show the following relation holds $$\#_B\circ \Phi=d\phi\circ \#_A?$$

Remarks.

$(i)$ $\Omega^k(A):=\Gamma(\Lambda^k A^*)$ for $k>0$ and $\Omega^0(A):=C^\infty(M)$ (that is, $\Omega^k(A)$ are the smooth sections of the vector bundle $\Lambda^k A^*$);

$(ii)$ The induced map $\Phi^*$ is given by $$(\Phi^* \omega)_p(v_1, \ldots, v_k)=\omega_{\phi(p)}(\Phi(v_1), \ldots, \Phi(v_k)).$$

$(iii)$ The exterior derivative $d_A$ is the map $d_A:\Omega^k(A)\longrightarrow \Omega^{k+1}(A)$ given by:

\begin{align*} \displaystyle (d_A\omega)(\alpha_0, \ldots, \alpha_k)&:=\sum_{j=0}^k (-1)^j \mathcal{L}_{\#_A\alpha_j}(\omega(\alpha_0, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k))\\ &+\sum_{i<j} (-1)^{i+j} \omega([\alpha_i, \alpha_j], \alpha_0, \ldots, \widehat{\alpha_i}, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k). \end{align*} Here we're using the isomorphism of $C^\infty(M)$-modules $$\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M)),$$ to define $d_A$. For degree zero: $$d_A(f)(\alpha):=\mathcal{L}_{\#_A \alpha}(f).$$

$(iv)$ The isomorphism $\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M))$ is given by: $$\omega\longmapsto ((\alpha_1, \ldots, \alpha_k)\longmapsto \omega(\alpha_1, \ldots, \alpha_k)),$$ where $\omega(\alpha_1, \ldots, \alpha_k):M\longrightarrow \mathbb R$ is the map: $$\omega(\alpha_1, \ldots, \alpha_k)(p):=\omega_p(\alpha_1(p), \ldots, \alpha_k(p)).$$ The inverse is given by: $$\theta\longmapsto (p\longmapsto \theta_p),$$ where $\theta_p:A_p\times \ldots\times A_p\longrightarrow \mathbb R$ is given by $$\theta_p(v_1, \ldots, v_k):=\theta(\alpha_1, \ldots, \alpha_k)(p),$$ where $\alpha_j\in \Gamma(A)$ are sections such that $\alpha_j(p)=v_j$ (which vary with $p$).

$(v)$ On the level of $\textrm{Alt}^k$ we could see $\Phi^*$ as given by:

\begin{align*} \displaystyle (\Phi^* \theta)(\alpha_1, \ldots, \alpha_k)(p)&=\theta(\beta_1, \ldots, \beta_k)(\phi(p)), \end{align*}

where $\beta_j\in \Gamma(B)$ are such that $\beta_j(\phi(p))=\Phi(\alpha_j(p))$.

Thanks.

Let $A\longrightarrow M$ and $B\longrightarrow N$ be Lie algebroids with anchors $\#_A$ and $\#_B$, respectively.

A morphism of Lie algebroids is a morphism of vector bundles $\Phi:A\longrightarrow B$ covering let's say $\phi:M\longrightarrow N$ such that the induced map $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$ commutes with the exterior derivative. How can I show the following relation holds $$\#_B\circ \Phi=d\phi\circ \#_A?$$

Remarks.

$(i)$ $\Omega^k(A):=\Gamma(\Lambda^k A^*)$ for $k>0$ and $\Omega^0(A):=C^\infty(M)$ (that is, $\Omega^k(A)$ are the smooth sections of the vector bundle $\Lambda^k A^*$);

$(ii)$ The induced map $\Phi^*$ is given by $$(\Phi^* \omega)_p(v_1, \ldots, v_k)=\omega_{\phi(p)}(\Phi(v_1), \ldots, \Phi(v_k)).$$

$(iii)$ The exterior derivative $d_A$ is the map $d_A:\Omega^k(A)\longrightarrow \Omega^{k+1}(A)$ given by:

\begin{align*} \displaystyle (d_A\omega)(\alpha_0, \ldots, \alpha_k)&:=\sum_{j=0}^k (-1)^j \mathcal{L}_{\#_A\alpha_j}(\omega(\alpha_0, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k))\\ &+\sum_{i<j} (-1)^{i+j} \omega([\alpha_i, \alpha_j], \alpha_0, \ldots, \widehat{\alpha_i}, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k). \end{align*} Here we're using the isomorphism of $C^\infty(M)$-modules $$\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M)),$$ to define $d_A$. For degree zero: $$d_A(f)(\alpha):=\mathcal{L}_{\#_A \alpha}(f).$$

$(iv)$ The isomorphism $\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M))$ is given by: $$\omega\longmapsto ((\alpha_1, \ldots, \alpha_k)\longmapsto \omega(\alpha_1, \ldots, \alpha_k)),$$ where $\omega(\alpha_1, \ldots, \alpha_k):M\longrightarrow \mathbb R$ is the map: $$\omega(\alpha_1, \ldots, \alpha_k)(p):=\omega_p(\alpha_1(p), \ldots, \alpha_k(p)).$$ The inverse is given by: $$\theta\longmapsto (p\longmapsto \theta_p),$$ where $\theta_p:A_p\times \ldots\times A_p\longrightarrow \mathbb R$ is given by $$\theta_p(v_1, \ldots, v_k):=\theta(\alpha_1, \ldots, \alpha_k)(p),$$ where $\alpha_j\in \Gamma(A)$ are sections such that $\alpha_j(p)=v_j$ (which vary with $p$).

$(v)$ On the level of $\textrm{Alt}^k$ we could see $\Phi^*$ as given by:

\begin{align*} \displaystyle (\Phi^* \theta)(\alpha_1, \ldots, \alpha_k)(p)&=\theta(\beta_1, \ldots, \beta_k)(\phi(p)), \end{align*}

where $\beta_j\in \Gamma(B)$ are such that $\beta_j(\phi(p))=\Phi(\alpha_j(p))$.

Thanks.

Let $A\longrightarrow M$ and $B\longrightarrow N$ be Lie algebroids with anchors $\#_A$ and $\#_B$, respectively.

A morphism of Lie algebroids is a morphism of vector bundles $\Phi:A\longrightarrow B$ covering let's say $\phi:M\longrightarrow N$ such that the induced map $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$ commutes with the exterior derivative. How can I show the following relation holds $$\#_B\circ \Phi=d\phi\circ \#_A?$$

Remarks.

$(i)$ $\Omega^k(A):=\Gamma(\Lambda^k A^*)$ (that is, $\Omega^k(A)$ are the smooth sections of the vector bundle $\Lambda^k A^*$);

$(ii)$ The induced map $\Phi^*$ is given by $$(\Phi^* \omega)_p(v_1, \ldots, v_k)=\omega_{\phi(p)}(\Phi(v_1), \ldots, \Phi(v_k)).$$

$(iii)$ The exterior derivative $d_A$ is the map $d_A:\Omega^k(A)\longrightarrow \Omega^{k+1}(A)$ given by:

\begin{align*} \displaystyle (d_A\omega)(\alpha_0, \ldots, \alpha_k)&:=\sum_{j=0}^k (-1)^j \mathcal{L}_{\#_A\alpha_j}(\omega(\alpha_0, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k))\\ &+\sum_{i<j} (-1)^{i+j} \omega([\alpha_i, \alpha_j], \alpha_0, \ldots, \widehat{\alpha_i}, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k). \end{align*} Here we're using the isomorphism of $C^\infty(M)$-modules $$\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M)),$$ to define $d_A$.

$(iv)$ The isomorphism $\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M))$ is given by: $$\omega\longmapsto ((\alpha_1, \ldots, \alpha_k)\longmapsto \omega(\alpha_1, \ldots, \alpha_k)),$$ where $\omega(\alpha_1, \ldots, \alpha_k):M\longrightarrow \mathbb R$ is the map: $$\omega(\alpha_1, \ldots, \alpha_k)(p):=\omega_p(\alpha_1(p), \ldots, \alpha_k(p)).$$ The inverse is given by: $$\theta\longmapsto (p\longmapsto \theta_p),$$ where $\theta_p:A_p\times \ldots\times A_p\longrightarrow \mathbb R$ is given by $$\theta_p(v_1, \ldots, v_k):=\theta(\alpha_1, \ldots, \alpha_k)(p),$$ where $\alpha_j\in \Gamma(A)$ are sections such that $\alpha_j(p)=v_j$ (which vary with $p$).

$(v)$ On the level of $\textrm{Alt}^k$ we could see $\Phi^*$ as given by:

\begin{align*} \displaystyle (\Phi^* \theta)(\alpha_1, \ldots, \alpha_k)(p)&=\theta(\beta_1, \ldots, \beta_k)(\phi(p)), \end{align*}

where $\beta_j\in \Gamma(B)$ are such that $\beta_j(\phi(p))=\Phi(\alpha_j(p))$.

Thanks.

Let $A\longrightarrow M$ and $B\longrightarrow N$ be Lie algebroids with anchors $\#_A$ and $\#_B$, respectively.

A morphism of Lie algebroids is a morphism of vector bundles $\Phi:A\longrightarrow B$ covering let's say $\phi:M\longrightarrow N$ such that the induced map $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$ commutes with the exterior derivative. How can I show the following relation holds $$\#_B\circ \Phi=d\phi\circ \#_A?$$

Remarks.

$(i)$ $\Omega^k(A):=\Gamma(\Lambda^k A^*)$ for $k>0$ and $\Omega^0(A):=C^\infty(M)$ (that is, $\Omega^k(A)$ are the smooth sections of the vector bundle $\Lambda^k A^*$);

$(ii)$ The induced map $\Phi^*$ is given by $$(\Phi^* \omega)_p(v_1, \ldots, v_k)=\omega_{\phi(p)}(\Phi(v_1), \ldots, \Phi(v_k)).$$

$(iii)$ The exterior derivative $d_A$ is the map $d_A:\Omega^k(A)\longrightarrow \Omega^{k+1}(A)$ given by:

\begin{align*} \displaystyle (d_A\omega)(\alpha_0, \ldots, \alpha_k)&:=\sum_{j=0}^k (-1)^j \mathcal{L}_{\#_A\alpha_j}(\omega(\alpha_0, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k))\\ &+\sum_{i<j} (-1)^{i+j} \omega([\alpha_i, \alpha_j], \alpha_0, \ldots, \widehat{\alpha_i}, \ldots, \widehat{\alpha_j}, \ldots, \alpha_k). \end{align*} Here we're using the isomorphism of $C^\infty(M)$-modules $$\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M)),$$ to define $d_A$. For degree zero: $$d_A(f)(\alpha):=\mathcal{L}_{\#_A \alpha}(f).$$

$(iv)$ The isomorphism $\Omega^k(A)\simeq \textrm{Alt}^k_{C^\infty(M)}(\Gamma(A), C^\infty(M))$ is given by: $$\omega\longmapsto ((\alpha_1, \ldots, \alpha_k)\longmapsto \omega(\alpha_1, \ldots, \alpha_k)),$$ where $\omega(\alpha_1, \ldots, \alpha_k):M\longrightarrow \mathbb R$ is the map: $$\omega(\alpha_1, \ldots, \alpha_k)(p):=\omega_p(\alpha_1(p), \ldots, \alpha_k(p)).$$ The inverse is given by: $$\theta\longmapsto (p\longmapsto \theta_p),$$ where $\theta_p:A_p\times \ldots\times A_p\longrightarrow \mathbb R$ is given by $$\theta_p(v_1, \ldots, v_k):=\theta(\alpha_1, \ldots, \alpha_k)(p),$$ where $\alpha_j\in \Gamma(A)$ are sections such that $\alpha_j(p)=v_j$ (which vary with $p$).

$(v)$ On the level of $\textrm{Alt}^k$ we could see $\Phi^*$ as given by:

\begin{align*} \displaystyle (\Phi^* \theta)(\alpha_1, \ldots, \alpha_k)(p)&=\theta(\beta_1, \ldots, \beta_k)(\phi(p)), \end{align*}

where $\beta_j\in \Gamma(B)$ are such that $\beta_j(\phi(p))=\Phi(\alpha_j(p))$.

Thanks.