The question I have is the following:


Let $Y,Y'$ be two integer homology 3-spheres. Can we embed $Y'$ into $Y\times I$ such that $Y'$ separates the two boundary components apart? 
Do we know any nontrivial examples of this type? (For example for $Y'=S^{3}$ or Brieskorn spheres?)


Also, we can consider the similar problem for knot concordance: are there two knots $K,K'$ with concordance $K\rightarrow K'$ and $K'\rightarrow K$ such that if we compose them, we get trivial concordance $K\rightarrow K$? Anyone can tell me examples of this kind? (especially for $K'$ a torus knot). 



What are known examples of a 3-manifold $Y$ embedded into $Y'\times I$ where $Y'$ is another 3-manifold?
To expand on Ian's answer to the first question: I proved many years ago (Seifert surfaces of knots in $S^4$, Pac. J. Math. 145 (1990), 97–116) that for any 3-manifold Y, there is a hyperbolic 3-manifold $Y'$ embedded in $Y \times I$ separating the boundary components, so that both pieces are homology cobordisms. This gives many non-trivial examples related to the first question. [Budney and Burton](http://arxiv.org/abs/0810.2346) have an excellent survey of the state of the art on embedding 3-manifolds in $S^4$, which explains most of the known obstructions, and exhibits lots of interesting examples. Note that it's much harder to specify $Y'$ and find $Y$; I would bet that (maybe assuming the Schoenflies conjecture) a `generic' $Y'$ can't be embedded in this fashion. 


The construction is related to the second question, which is in part answered by the existence of doubly-slice knots, ie the special case when $K$ is the unknot. Examples can be found in Fox's Quick Trip through Knot Theory, and the subject was expanded quite a bit by De Witt Sumners (Invertible knot cobordisms. Comment. Math. Helv. 46 1971 240–256). There's a fair amount of literature on the subject, mostly giving obstructions to a given knot $K'$ being a slice of $\mathcal{O} \times I$ (i.e. $K'$ being doubly slice). I would make the same bet as above (without any assumptions) in the knot-theory setting: for a `generic' $K'$, there is no $K$ with $K'$ splitting $K \times I$.


