Homotoping curves in 3/4-manifolds to lie on embedded tori? I am wondering if every simple closed curve $c$ in a 3-manifold $Y$ can be homotoped to lie on an embedded torus?  Maybe I need to assume that the 3-manifold is orientable?
[EDIT: As @JohnHughes points out - the answer is yes if we assume $Y$ is orientable, or more generally if $c$ is orientation preserving.  ]
Similarly, I am wondering if every simple closed curve $c$ in a 4-manifold $X$ can be homotoped to lie on an embedded torus?
Any sufficient conditions on $X$ are also appreciated if the answers to the above turn out to be no. 
 A: If $Y$ is orientable and smooth and compact, I think it's true:
Take your curve $\gamma$; the tubular neighborhood theorem says that a neighborhood of $\gamma$ looks like a 2-disk-bundle over $\gamma$, and since these are classified by "clutching functions" in $\pi_0(O(2))$, there are only two possibilities. For the case where the clutching element is in $O(2)- SO(2)$, you've got a nonorientable bundle, and that makes $\gamma$ an orientation-reversing look in $Y$, which contradicts the "orientable" assumption. So...it's actually diffeomorphic to $S^1 \times D^2$. Now consider the image of $S^1 \times C$, where $C$ is the radius-one-half circle in $D^2$ under this diffeomorphism. That'll be a torus in $Y$. And picking a section of the resulting $S^1$ bundle over $\gamma$ gives a curve homotopic to $\gamma$ that lies on the surface of the torus. 
@Camilo's answer seems to be addressing the harder question of whether it lies on an embedded torus that has some non-triviality property that I cannot quite discern. 
