Breaking length is bounded by the index drop

The components of the given broken trajectory are nonconstant and run from $p_{i-1}$ to $p_i$, so [F1] applies to the string $p=p_0\to p_1\to \cdots \to p_r=q$ and gives $\lambda (p_0)>\lambda (p_1)>\cdots >\lambda (p_r)$; each difference $\lambda (p_{i-1})-\lambda (p_i)$ is a positive integer by [F5], hence at least one.

Telescoping the $r$ drops gives $\lambda (p)-\lambda (q)={\sum }_{i=1}^r(\lambda (p_{i-1})-\lambda (p_i))\ge r$, hence $r\le \lambda (p)-\lambda (q)$.

Each broken trajectory determines its length $r$, its string of critical points $p=p_0,\dots ,p_r=q$ and its tuple $({\gamma }_1,\dots ,{\gamma }_r)\in \mathcal{M}(p_0,p_1)\times \cdots \times \mathcal{M}(p_{r-1},p_r)$, and no two different data give the same broken trajectory by [F4]; conversely a tuple whose string satisfies $\lambda (p_{i-1})>\lambda (p_i)$ for every $i$ yields a broken trajectory, because the components are then nonconstant. Therefore $\bar{\mathcal{M}}(p,q)$ is the disjoint union of the products over $r\ge 1$ and over such strings. Only strictly index-decreasing strings are included, so consecutive points are distinct and every moduli-space factor is defined.

If $\lambda (p)-\lambda (q)=2$ and a broken trajectory has length $r\ge 2$, then step 2.1 gives $r\le 2$, so $r=2$: the trajectory is once-broken and has exactly one intermediate critical point $p_1$. Its two drops are positive integers with sum $2$ by step 1.1, hence both equal $1$, that is $\lambda (p_1)=\lambda (p)-1$.

The union is finite: by step 2.1 only the integers $1\le r\le \lambda (p)-\lambda (q)$ occur (and $\bar{\mathcal{M}}(p,q)=\mathrm{\varnothing }$ when $\lambda (p)\le \lambda (q)$), and for each such $r$ the string is a finite sequence of critical points chosen from the finite set $\mathrm{Crit}(f)$ by [F2]; hence finitely many products occur, each contributing as a single term of the disjoint union irrespective of its cardinality.