TITLE: Does the Morse homology depend on the orientation?
QUESTION [5 upvotes]: Before asking my question I need to define some objects. I will follow the book "M. Audin, M.Damian - Morse theory and Floer homology", but the terminology is quite standard:
Let $M$ be a smooth compact manifold and consider a Morse-Smale pair $(X,f)$ on $M$ ($X$ is a gradient-like vector field and $f$ is an adapted Morse function). If $a,b$ are two critical points of $f$, we indicate with $\mathcal L(a,b)$ the manifold such that every point is a trajectory of $X$ ''starting'' from $a$ and ''ending'' in $b$. One can show that if $\text{ind}(a)=\text{ind}(b)+1$ then $\mathcal{L}(a,b)$ is a finite set. Moreover if we orient the stable manifold  $W^s(a)$, remember that it is a disk, we induce an orientation on $\mathcal L(a,b)$, namely at each point we associate $\pm 1$ if $\text{ind}(a)=\text{ind}(b)+1$.
At this point one can define the Morse-Smale complex:
$$C_k:=\sum_{a\in\text{Crit}_k(f)}\mathbb Za$$
where clearly $\text{Crit}_k(f)$ is the set of critical points of index $k$. The map $d_k:C_k\longrightarrow C_{k-1}$ acts on the generators of $C_k$ in the following way:
$$d_k(a)=\sum_{a\in\text{Crit}_{k-1}(f)}N(a,b)b$$
where $N(a,b)\in\mathbb Z$ is the sum of the $\pm 1$ (the orientations) attached to the points of $\mathcal L(a,b)$.

Question: From the above construction it is evident that the Morse-Smale complex (in particular the number $N(a,b)$) depends on the
  orientation that we fix on the stable manifolds $W^s(a)$. This sounds
  very strange to me, indeed I'd expect a complex independent from the
  orientation. Maybe by passing to the homology group one can recover
  the independence but I can't see it.

Many thanks.

REPLY [2 votes]: If you change the orientation on $W^s(a)$ then an isomorphism from the old complex to the new is given by sending the generator $a$ of $C_k$ to $-a$ (and acting as the identity on all other critical points).