Every contractible smooth loop has a neighbourhood with $H^2=0$

Question

Let $c: S^1 \to M$ be a smooth contractible loop (not necesarily an embedding, or even an immersion) on the connected, compact symplectic manifold $(M,\omega)$ (if this helps somehow, $c$ is a $1$-periodic orbit of a $1$-periodic time-dependent Hamiltonian $H_t$). The book by Audin, Damian claims that there exists a neighbourhood $U$ of $c(S^1)$ that retracts (and by this, they mean "deformation retracts") onto $c(S^1)$.

What is needed, however, is the fact that there exists a neighbourhood $U$ of $c(S^1)$ such that $H^2(U;\mathbb{R})=0$. My question is: why is this true?

If $\dim(M)=2$, then this follows by considering $p \notin c(S^1)$ and taking $U:=M\backslash\{p\}$. Since $U$ is a non-compact connected $2$-manifold, it follows that $H^2(U;\mathbb{R})=0$. However, I can't see how to proceed in generality, since tubular neighbourhoods (which I think is what the authors had in mind) seem elusive etc.

Answer 1

Inductively isotope the skeleta of a triangulation of $M$ so that the $k$-skeleton is transverse to $c$. (We may use an isotopy to achieve transversality because the simplices are already embedded, and we don't need to homotope the boundary at all; a small homotopy of only the interior of the simplex doesn't change the property of being an embedding. Demand that a neighborhood of the rest of the triangulation is kept fixed, then apply isotopy extension to move the whole mess around.) Once this is true of all skeleta, $\text{Im } c \cap \sigma$ is empty unless $\sigma$ is top dimensional, or one dimension below that. If we consider the union of the open faces of all $n$ and $(n-1)$-dimensional simplices of $M$, we have an open manifold $M_\tau$ with the homotopy type of a graph (the graph whose vertices correspond to $n$-simplices and edges correspond to $(n-1)$-simplices) with $\text{Im } c \subset M_\tau$. In particular, $\text{Im } c$ has a neighborhood with $H^2(M_\tau) = 0$.