I am currently learning Morse theory through Milnor's book on the subject. I am trying to understand the result given in p.96 that tells us the cell decomposition of the loop space of $S^n$:
Corollary 17.4. The loop space $\Omega(S^n)$ has the homotopy type of a CW-complex with one cell each in the dimensions $0$, $n-1$, $2(n-1)$, $\dots$
The computation is done by fixing two basepoints $p, q \in S^n$ and counting the index of the geodesics from $p$ to $q$. The explanation given in p.95-96 is perfectly sound, but the use of the word loop space is bugging me.
To apply the fundamental theorem of Morse theory (p.95, Theorem 17.3) to the path space $\Omega(S^n; p, q)$, we need the assumption that $p$ and $q$ are not conjugate along geodesics, i.e. $q \neq \pm p$ in this case. The counting of indices of geodesics in p.96 also uses this assumption.
The loop space, on the other hand, seems to indicate the case $p=q$, i.e. $\Omega(S^n) = \Omega(S^n; p, p)$. Hence my question is:
Why does the computation in p.96, assuming $q \neq \pm p$, still tells us about the loop space of S^n? Is it the case that if $q$ is a non-conjugate point to $p$ that is "close enough" to $p$ we can say $\Omega(S^n; p, q) \simeq \Omega(S^n; p, p)$? If that is the case, what would be the homotopy equivalence?
In general, if $X$ is any path-connected space and $a,b,c,d\in X$, then the path spaces $\Omega(X;a,b)$ and $\Omega(X;c,d)$ are homotopy equivalent. Indeed, if you choose a path from $a$ to $c$ and a path from $d$ to $b$, concatenating with these paths gives a map $\Omega(X;a,b)\to\Omega(X;c,d)$ and concantenating with the inverse paths gives a map $\Omega(X;c,d)\to\Omega(X;a,b)$. The composites are homotopic to the identity: for instance, the composite $\Omega(X;a,b)\to \Omega(X;a,b)$ is given by concatenating with nullhomotopic loops at $a$ and $b$ (the paths we chose together with their inverses), and so you can use nullhomotopies of these loops to get a homotopy to the identity map.