Prove Poincare duality theorem with Morse theory. First let us consider a smooth n-manifold. And find a Morse function $f$. Now let's consider $-f$. A singular point of $f$ with index $k$ is a singular point of -f with index n-k. Thus we have a canonical one-one correspondence between $C_k(M)$ and $C^{n-k}(M)$ where I'm considering the cellular chain and cochain groups. My question is can I deduce the Poincare duality theorem by analyzing carefully the behavior of boundary and co-boundary maps? But I don't see where is the condition orientable needed.
 A: The Morse complex on a manifold (without orientation assumptions) can be defined using the following data: A Morse function $f$, a metric $g$ and a choice of orientation $\mathfrak{o}$ of the unstable manifolds. Then the space $W(x,y)=W^u(x)\cap W^s(y)$ is given an orientation by the following exact sequence
$$
0\rightarrow TW(x,y)\rightarrow TW^u(x)\rightarrow NW^s(y)\rightarrow 0.
$$
The two spaces on the right are oriented by the choice $\mathfrak{o}$, as $NW^s(y)\cong TW^u(y)$. If two of the three vector bundles in an exact sequence are oriented, so is the third.
This is a general phenomenon (see for example Hirsch' book): the transverse intersection of a oriented and a cooriented submanifold is a oriented submanifold.
Now to define the dual complex one needs an orientation of the stable manifolds. Note that $\mathfrak{o}$ doesn't give this, it gives a coorientation. However, in an oriented manifold we have the exact sequence of oriented vector bundles
$$
0\rightarrow TW^u(x)\rightarrow TM\rightarrow TW^s(x)\rightarrow 0.
$$
Thus the stable manifolds are oriented, and we can do the count for $-f$. The orientation of $W(x,y,f)$ and $W(y,x,-f)$ need not agree, but if the manifold is oriented, the orientation difference depends only on the degree of $x,y$ and the dimension of the manifold. Thus we get a nice isomorphism.
It is good to work this out in the case where it fails, i.e. your favourite function on $\mathbb{R}\mathbb{P}^2$.
