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Aug 20, 2012 at 12:27 comment added Patricia Hersh Vel, my recollection is that Robin Forman made this statement various times at conferences that a discrete Morse function implies a simple homotopy equivalence, so probably that's how it became folklore. I think it's true, and hopefully the argument I gave above explains why.
Aug 20, 2012 at 2:35 history made wiki Post Made Community Wiki by S. Carnahan
Aug 19, 2012 at 18:49 comment added Patricia Hersh @Vel: I think one can also handle the critical cells by using some anticollapses, but I don't know a reference for this. Idea: once one removes critical cell $C$, one can see what elementary collapses to do to get down to $X^a$ and how they would carry the boundary of $C$ to have new attaching map $f_{C_a}$. Therefore, we first do an anticollapse by adding in cell $C'$ with attaching map $f_{C_a}$ along with a cell $D$ of dimension one higher that has $C'$ as a free face and also attaches to the cells $\sigma $ with $a\le f(\sigma ) \le b$. Now collapse away $C,D$ and the noncritical cells.
Aug 19, 2012 at 17:50 comment added Vidit Nanda We have to be careful with the simple homotopy claim. Given $f:X \to \mathbb{R}$ and setting $X^a = \lbrace \sigma \in X~|~f(\sigma) < a\rbrace $ there is a simple homotopy equivalence between $X^a$ and $X^b$ provided there are no critical values in $(a,b)$. On the other hand, when we cross a critical value, then we only have homotopy equivalence coming from the attaching map of the boundary of the critical cell: this need not be a simple homotopy equivalence.
Aug 19, 2012 at 17:46 comment added Patricia Hersh Thanks! My description of the analogy was for Forman's notion, but it's a good idea to add this reference.
Aug 19, 2012 at 16:02 comment added Lee Mosher You might add as a reference the paper of Bestvina and Brady, Morse theory and finiteness properties of groups. Invent. Math. 129 (1997), no. 3, 445–470.
Aug 19, 2012 at 15:52 history answered Patricia Hersh CC BY-SA 3.0