Suppose we have a Hamiltonian action of a torus $T = T^m = R^m/Z^m$ on a compact, connected symplectic manifold $M$. According to the convexity theorem, we know every fiber of the momentum map $\mu: M\to R^m$ is connected. My question here is about the proof.
We assume the Hamiltonian action is effective without loss of generality, i.e., only the zero point of $T$ fixes $M$. I already know that the set of regular values of $\mu$ is dense in $\mu(M)$, also the set of points $\eta$ in $\mu(M)$ with $(\eta_1, \dotsc , \eta_{m-1})$ a regular value for the reduced momentum map $(\mu_1, ..., \mu_{m-1})$ is dense in $\mu(M)$. I also know that the fiber of $\eta$ is connected whenever $(\eta_1, ..., \eta_{m-1})$ a regular value for the reduced momentum map. "Since the set of such points is dense in $\mu(M)$ it follows by continuity that the fiber of $\eta$ is connected for every regular value $\eta$." (in the book by McDuff and Salamon), and this is my first question. My second question is that then we can imply that every fiber of $\mu$ is connected?
Notice that here the Hamiltonian action must play a special role, as the following is not true:
Suppose $f: M \to N$ is a smooth map between two smooth manifolds, with $M$ compact and connected, and suppose there is a dense subset of $f(M)$ where each fiber is connected, then each fiber of $f$ is connected. Example: consider a natural smooth surjection from $S^1$ to the figure eight. The fiber over the nodes of the figure eight has two points but every other fiber is a single point.
If anyone knows how to prove the connectedness part of the Convexity Theorem, could you please show us? Thank you very much!