There are more general definitions, but for my purposes a **Lie algebroid** on a smooth manifold $X$ is a vector bundle $A \to X$, a map $\rho: A \to {\rm T}X$ of vector bundles over $X$, and a bracket $[,]$ making $\Gamma(A)$ into an $\mathbb R$-Lie algebra, such that $\rho$ induces is a map $\Gamma(A) \to \Gamma({\rm T}X)$ of Lie algebras and such that the Leibniz rule $[a,fb] = f[a,b] + (\rho(a)f)b$ is satisfied for $f\in \mathcal C^\infty$ and $a,b\in \Gamma(A)$.

So, suppose I happen to have a Lie algebroid $A\to X$ lying around, and also a smooth map $\phi: Y\to X$. (In my case, $Y \to X$ happens to be a vector bundle, so you can assume some fairly strong properties of the map.) Then I can certainly pull back the vector bundle $A\to X$ to $\phi^*A \to Y$. Is there a natural Lie algebroid structure I can put on this pullback? The answer is probably "how natural do you want it?": if $Y = \{{\rm pt}\}$ and $\phi({\rm pt}) = y\in X$, then the anchor map for $\phi^*A \to \{{\rm pt}\}$ must be trivial, but the only part of the fiber $A_y$ with a Lie algebra structure is $\ker \{A_y \overset\rho\to {\rm T}_y X\}$.

So the real question is:

Along what type of maps do Lie algebroids pull back? In particular, can I always pull back a Lie algebroid along a submersion?