A coauthor of mine and I want to use the following innocent looking statement in a forthcoming paper:
Statement. Let $M^{2n}$ be a compact manifold and let $f$ be a Morse function with critical points of even indices. Then for some choice of a metric $g$ on $M^{2n}$, the closure of each unstable manifold in $M^{2n}$ is a cycle of dimension equal to the index of the corresponding critical point.
I thought naively that such a statement should be contained in some classical book (but the answers given below indicate that this might be not the case).
Here is an idea of how to deduce the statement from the literature. Let's take the paper of Francois Laudenbach
http://www.numdam.org/article/AST_1992__205__219_0.pdf
and look into Remark 3. This remark claims something much stronger, namely that even without assumption on even indices the union of unstable manifolds give a structure of a CW complex on $M$ in case there exists a metric $g$ on $M$ such that the gradient flow satisfies the Morse-Smale condition and additionally the gradient vector field is Special Morse (i.e. looks like $\sum_i{\pm}x_i\frac{\partial}{\partial x_i}$).
Unfortunately, it is not stated in this paper whether such a metric $g$ always exists (added: according to John and Alessia this is very simple)
Question. Is there a reference or short proof for the above Statement? Or maybe one can say that a metric satisfying Morse-Smale condition and Special Morse condition always exists?
Added. I would like to thank John, Pietro and Alesia for answers. I still hope that the exact Statement that I want might be from 20th century, not 21st. Indeed, suppose that all the indices are even, and $g$ is Morse-Smale. Then for each unstable cell $W$ the set $\bar W\setminus W$ has Hausdorf dimension at most $\dim W-2$. Should not this give a well-defined cycle in $M^{2n}$?
Question 2 I don't quite understand what is Morse Homology, but should not the above Statement be a trivial part of this theory?
(what about this preprint: https://arxiv.org/pdf/math/9905152.pdf ? looks relevant)