I'm trying to understand the one-shot deviation principle in the context of an extensive game. It appears to me that the payoff $\pi_i()$ for each player $i$ in each stage $s$ must be independent of the path taken to reach $s$. Otherwise, it is easy to come up with a counter example. For example, say $\pi_1(AA)=\pi_1(AB)=\pi_1(BA)=0, \pi_1(BB)=3$. Here, $AA$ is unimprovable for player 1, but is clearly not optimal, because $BB$ is better.
However, in an extensive game, the strategies for the different stages are not independent, making it difficult to write down a game tree with independent payoffs in the different stages. For example, with the Prisoner-Revenge game (p.176 in this book), the first stage is a standard Prisoner's dilemma with actions Mum (M) and Fink (F), and the second stage is either Loner (L) or Gang (G) with two Nash equilibria at (L,L) and (G,G). This results in a strategy set of $2^5=32$ strategies for each player, corresponding to {M,F} $\times$ {L|MM,G|MM} $\times$ {L|MF,G|MF} $\times$ {L|FM,G|FM} $\times$ {L|FF,G|FF}. Now, if we write these 32 strategies as a $2\times 2\times 2\times 2\times 2$ game tree for Player 1 and fix Player 2's strategy to be, say, F, G|MM, L|MF, L|FM, G|FF (shortened to FGLLG), then we have that Player 1's payoff for (L/G)|MF and (L/G)|FF are different, depending on whether his first stage action was M or F.
To make it clearer, let's say in the first stage, Player 1 plays M. Now since the outcome in the first stage is MF, his payoff only depends on his strategy in playing L|MF vs G|MF. His other strategies given MM, FM, or FF are irrelevant. On the other hand, if he plays F in the first stage, then only his strategy in (L/G)|FF is relevant.
How can we represent this game in a game tree such that we can apply the one-shot deviation principle?