# Homotopy of algebroid paths

In Lectures on Integrability of Lie Brackets, Proposition 3.15, Fernandes and Crainic motivate what it means to say that two paths on an algebroid are homotopic. I spent quite some time trying to grasp the proof, but it still eludes me. I will write down some definitions as to make this question self-contained, but everything I write down can be looked up in their document.

• For an algebroid $$\pi : A \to M$$ with bracket $$[\cdot,\cdot]$$ and anchor $$\rho$$, they define the notion of an $$A$$-path as a $$C^1$$-path $$a : [0,1] \to A$$ such that $$\rho \circ a = \frac{d \gamma}{dt}$$, where $$\gamma = \pi \circ a : [0,1] \to M$$ is the corresponding base-path.

• A variation of $$A$$-paths is a family of $$A$$-paths $$(a_\epsilon)_{\epsilon \in [0,1]}$$ which is $$C^2$$ in $$\epsilon$$ and so that the end points of the base paths are fixed.

• Lastly, if $$\alpha(s,\cdot) \in \Gamma(A)$$, $$s \in [0,1]$$ are time-dependent sections, their flow is the the unique maximal family of $$A$$-automorphisms $$\{\phi^{t,s}_\alpha\}_{t,s}$$ with $$\phi^{t,s}_\alpha \phi^{s,u}_\alpha = \phi^{t,u}_\alpha$$, $$\phi^{t,t}_\alpha = \text{id}$$ and \begin{align*} \frac{d}{dt} \Big \vert_{t = s} \left(\phi^{t,s}_\alpha\right)^\star (\beta) = [\alpha^s,\beta] \quad \forall \beta \in \Gamma(A) \end{align*} with $$(\phi^{t,s}_\alpha)^\star (\beta)(x) = \phi^{s,t}_\alpha \beta(\phi^{t,s}_{\rho(\alpha)}(x))$$, where $$\phi^{t,s}_{\rho(\alpha)}$$ is the usual time-dependent flow on $$M$$.

With these definitions in place, in Proposition 3.15 they are given a variation of $$A$$-paths $$(a_\epsilon)$$ and a time-dependent section $$\xi_\epsilon$$ so that $$\xi_\epsilon(t,\gamma(t)) = a_\epsilon(t)$$. In the proof, they define $$\eta (\epsilon,t,x) := \int_0^t \phi^{t,s}_{\xi_\epsilon} \frac{d \xi_{\epsilon}}{d \epsilon} (x,\phi^{s,t}_{\rho(\xi_\epsilon)} (x)) ds \in A_x.$$

They prove that the so-defined section $$\eta(\epsilon,t,\cdot) \in \Gamma(A)$$ has the property $$\frac{d \eta}{d t } - \frac{d \xi_\epsilon}{d \epsilon} = [\eta,\xi].$$

Up to here, I think I have been able to follow the proof, but now I am lost: They define $$X := \rho(\xi), Y := \rho(\eta)$$ with the anchor of the algebroid, and say these fulfil a similar equation, presumably $$\frac{dY}{dt} - \frac{dX}{d\epsilon} = [Y,X],$$ simply by applying the anchor map.
They then claim that because $$X(\epsilon,t,\gamma_\epsilon(t)) = \rho( a_\epsilon(t)) = \frac{d \gamma_\epsilon}{d t}(t)$$ (true as property of an $$A$$-path), it follows that $$Y(\epsilon,t,\gamma_\epsilon(t)) = \frac{d \gamma_\epsilon}{d \epsilon}$$. I have no idea how that follows and have tried a couple different unsuccessful approaches. It seems like this has something to do with $$[Y,X](t,\epsilon,\gamma_\epsilon(t)) = 0$$, which may well be true, but I cannot see a good reason. Apologies for this long, convoluted and very specific question, but I am kinda hoping that there is an absolute algebroid-pro running around who can instantly see the solution to my troubles. Thank you!