TITLE: General position for map from surface to 3-manifold
QUESTION [6 upvotes]: Let f be a smooth map from a (compact,oriented) surface S to a (compact, oriented) 3-manifold M. Suppose that I have an embedded (non-contractible) loop $\gamma$ in my surface $S$, can I find an (immersed) loop $\gamma'$ freely homotopic to $f \circ \gamma$ which is disjoint from $im(S)$?

REPLY [9 votes]: In general, no you cannot. Consider one dimension down. Take two curves on a torus, intersecting transversely in a point. One of the curves cannot be homotoped to be disjoint from the pair of curves. Now, cross with a circle $S^1$ to get $T^3$. The pair of curves crossed with $S^1$ is an immersed surface (two immersed tori $T^2$). Then either curve lies in the surface, but may not be homotoped to be disjoint from the immersed surface (this may be proved using the intersection product on homology, dual to the cup product). 
You might object that the surface is disconnected. To obtain a connected surface, just tube them together to get a homologous surface.