Let $M$ be a closed connected 3-manifold. I know that every element of $H_2(M;\mathbb{Z})$ can be represented by an embedded surface. I am wondering if I can always find a generating set of disjoint embedded surfaces? Assuming the answer is no, are there some natural sufficient conditions that all one to do this?
I have a feeling that I just missed a day of algebraic topology and can't think of some basic obstruction. I also have the issue of not knowing too many 3-manifolds.
The obstruction you're looking for is the statement that if $M_1, M_2 \subset N$ are closed oriented submanifolds of a closed oriented manifold, then $PD(M_1 \cap M_2) = PD(M_1) \smile PD(M_2)$. That is, Poincare Duality takes the cup product to the intersection product and vice versa. In particular, if the cup product $H^1(M;\Bbb Z) \otimes H^1(M;\Bbb Z) \to H^2(M;\Bbb Z)$ pairs different elements nontrivially together, then you cannot represent a basis of $H_2(M;\Bbb Z)$ by disjoint surfaces. The easiest example is probably $T^3$.