Index-one trajectory moduli spaces are finite

Let $(v_n)$ be a sequence in $\mathcal{M}(p,q)$. By [F2] a subsequence converges geometrically to a broken trajectory whose number of components is at most $1$; a broken trajectory has length at least one, so the limit is an ordinary trajectory, that is, an element of $\mathcal{M}(p,q)$. Hence every sequence in $\mathcal{M}(p,q)$ has a convergent subsequence: the space is sequentially compact.

By [F1] the space $\mathcal{M}(p,q)$ is a topological manifold, hence metrizable by [F3]; for metrizable spaces sequential compactness is equivalent to compactness, so step 1.1 makes $\mathcal{M}(p,q)$ compact. This is where the choice principles enter: [A1] supplies the full AC assumed by the metrization result, and $\mathrm{D}\mathrm{C}$ and ${\mathrm{A}\mathrm{C}}_{\omega }$ assumed by the metric compactness equivalence in [F3].

By [F1] the compact space $\mathcal{M}(p,q)$ is discrete, so by [F4] it is finite. For the second clause: whenever a differential on this page is defined by counting trajectories between critical points of Morse index drop one, each of its coefficients is the cardinality (modulo two) or the signed count of a space of the form $\mathcal{M}(p,q)$ with $\lambda (p)-\lambda (q)=1$, and each such space is finite by the first clause; the sums defining the differential therefore have only finitely many nonzero terms, and the total number of coefficients is finite because the critical set of a Morse function on a closed manifold is finite by [F5].