Reverse accumulation sums gradients over shared parameter uses

Since $T$ is finite, the initial map, $T$ transition evaluations, and scalar loss form a finite program whose one parameter input $\theta$ is used at each transition and at the loss. Each map is $C^1$ on an open domain containing the trajectory, so repeated application of [F4] makes $J$ differentiable there; on a fixed strict branch the same conclusion holds branchwise.

The algorithm's primal pass evaluates this unrolled program. Its reverse pass seeds the loss, visits transitions in descending order, and sends each transition's parameter pullback to the same $\theta$ input while sending its state pullback to $h_{t-1}$; the initial-state pullback is then added at $\theta$. This is the reverse sweep on the finite graph in [F1], so [F3] gives the exact gradient of $J$.

The terminal state's cotangent is ${\lambda }_T=(D_h\mathrm{\ell }{)}^T$. At transition $t$, the chain rule gives the cotangent at $h_{t-1}$ as $B_t^T{\lambda }_t$ and adds $A_t^T{\lambda }_t$ to the derivative with respect to the shared parameter while $h_{t-1}$ is held fixed. Descending induction therefore gives the stated ${\lambda }_{t-1}$ recurrence and sums one parameter contribution for each of the $T$ transition uses.

The loss's direct dependence on $\theta$ contributes $(D_{\theta }\mathrm{\ell }{)}^T$, and the dependence of $h_0(\theta )$ contributes $(D{h}_0{)}^T{\lambda }_0$. Adding these to the $T$ transition contributions from step 2.2 gives the displayed formula. If $T=0$, there are no transition terms and the same chain rule reduces $J(\theta )=\mathrm{\ell }(\theta ,h_0(\theta ))$ to the two remaining terms. This is exactly the algorithm's return value and gradient.