# How to show that the inverse image of a face of a simple polytope is a connected manifold?

If $M^{2n}$ is a toric manifold over a simple polytope $P^n$ i.e; the orbit space of the action of the $(S^1)^n$ on $M^{2n}$ is an $n$ dimensional simple polytope $P^n$. Let $\pi : M^{2n} \rightarrow P^n$ be the orbit map.

Here by a toric manifold of dimension $2n$ I mean $M^{2n}$ has an $(S^1)^n$ action with the following data : for every $x \in M^{2n}$ there are

(i) an automorphism $\theta_x : (S^1)^n \rightarrow (S^1)^n$

(ii) $(S^1)^n$ invariant open sets $V$ in $M^{2n}$ and $W$ in $\mathbb C ^n$ where $V$ contains $x$

(iii) a $\theta_x$ equivariant homeomorphism $f : V \rightarrow W$

where the $(S^1)^n$ action on $\mathbb C^n$ is just pointwise multiplication.

If $F^k$ is a $k$ dimensional face of $P^n$ then I need to show that $S = \pi ^{-1} (F^k)$ is a $2k$ dimensional connected submanifold of $M^{2n}$.

I am unsure how to proceed. If I just take the charts on $M^{2n}$ and restrict them to $S$ then will it work? If $(U_{\alpha},{\phi}_{\alpha})$ is a chart on $M^{2n}$ then will $\phi_{\alpha}(S \cap U_{\alpha})$ be an open subset of $\mathbb{C}^k$?

Also, to show connectedness I have to show that $\pi : \pi^{-1}(int(F^k)) \rightarrow int(F^k)$ is a trivial $(S^1)^k$ bundle. Where $int(F^k)$ is the relative interior of $F^k$. I am unsure how to do this as well. But I think the contractibility of $int(F^k)$ should come into play here.

Thanks!