TITLE: Unique factorization of manifolds?
QUESTION [6 upvotes]: I wonder if there is a result on the unique factorization of manifolds.
Call a topological manifold to be indecomposable if it is not homeomorphic to a product of manifolds of positive dimension. Is every manifold a unique (up to order) product of indecomposable ones?
I couldn't find any statements on this simple question. Are there any results on this? Any result in different categories (smooth, complex, Riemannian or whatever) or with extra conditions is fine.
[edit]
The answer seems to be No in most cases. Can we impose strong conditions so that the answer is positive?

REPLY [7 votes]: Generally the answer is no.  For example, $TS^2$ is indecomposible.  But $TS^2 \times \mathbb R \simeq S^2 \times \mathbb R^3$, so $S^2 \times \mathbb R^3$ splits as a product of indecomposibles in several different ways.  
You could use $\mathbb C$ instead of $\mathbb R$ if you want complex manifolds.  
You get similar things happening for Riemann manifolds as well.