Let $A\longrightarrow M$ and $B\longrightarrow N$ be [Lie algebroids](https://en.wikipedia.org/wiki/Lie_algebroid) 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.



Property of Lie algebroid morphism: $\#_B\circ \Phi=d\phi\circ \#_A$?
A vector is determined by its action on functions as a directional derivative. This is just the Lie derivative in degree $0$ mentioned at the end of $(iii)$. Together with the naturality $(ii)$ of $d$ and the definition of $d\phi$, you get $$(\#\_b\circ\Phi\_p)(\alpha)(f)=d\_B(f)\_{\phi(p)}(\Phi\alpha)=(\Phi^\*d\_Bf)\_p(\alpha)=(d\_A(f\circ\phi))\_p(\alpha)=(d\phi\_p\circ\#\_A)(\alpha)(f)$$
for all $f\in C^\infty(N)$ and all $\alpha\in A\_p$. Your claim follows.


