Homology class of limit of surfaces

Question

Let $(M,g)$ be a compact, connected and oriented Riemannian $3$-manifold with nonempty boundary. Suppose a sequence $\{S_n\}_{n \geq 1}$ of compact, connected, oriented and properly embedded surfaces (meaning that $\partial S_n = \partial M \cap S_n$) with nonempty boundary converges smoothly to a properly embedded surface $S$ with boundary. Is it true that there exists $N \geq 1$ such that $[S_n] = [S] \in H_2(M, \partial M; \mathbb{Z})$ for all $n \geq N$? Or is it true that if $[\Sigma_n]\neq 0$ for all $j \geq 1$ then $[\Sigma] \neq 0$?

Answer 1

Letting $N(S)$ be a regular neighborhood of $S$, we can apply the normal bundle theorem to identify $N(S)$ diffeomorphically with the normal bundle of $S$, so that under this identification the surface $S$ itself is identified with the zero section of the normal bundle.

Presumably by "smooth convergence" you mean convergence in the $C^1$ Hausdorff topology. That being the case, it follows that for sufficiently large $n$, say $n \ge N$, the surface $S_n$ is contained in the bundle $N(S)$ and is tranverse to the fibers of that bundle, intersecting each fiber in exactly one point. The fiber segments of the bundle that connect $S$ to $S_n$ may therefore be used to define a smooth isotopy $S \times [0,1] \to N(S)$ between $S_n$ and $S$.

From this it follows for all $n \ge N$ we have $[S_n] = \pm [S]$ in $H_2(M,\partial M;\mathbb Z)$. More than that cannot be said, because you have not specified any kind of compability information about orientations.