Assume we have an oriented manifold $M$ of dimension say $n > 2$ (or if necessary $n > 3$) and an embedded circle $S$ whose fundamental class $[S]$ is zero in homology. Is that circle always the boundary of an embedded disc?
In dimension 2, there is the counter-example taking a 2-genus surface and the circle splitting the 2-genus into two 1-genus surfaces with boundary. This seems to work here because of an Jordan-curve-like argument. I wonder whether this might change in higher dimensions.
Counter-examples or proof-sketches/references are pretty much appreciated.
The first obstruction is given by thinking of the circle $S$ as being an element of the fundamental group. A necessary requirement that $[S]$ bounds an embedded disk is that $[S]=0 \in \pi_1(M)$. This is very much going to be false in general.
If, however, you know that $[S]=0$ in the fundamental group, if the dimension of $M \geq 5$, then the answer is yes. This follows from the following form of the Whitney embedding theorem: If $X \rightarrow Y$ is a map of smooth manifolds such that $2\operatorname{dim}(X)+1 \leq \operatorname{dim}(Y)$, then it is homotopic to an embedding.
We apply this theorem to the map $D^2 \rightarrow M$ guaranteed by the fact $[S]=0$ in the fundamental group to get the embedded disk you ask for.