# help me in trace of following proposition

In a paper an author proved the following proposition

Proposition: let $f$ be a homeomorphism of a connected topological manifold $M$ with fixed point set $F$. then either $(1)$ $f$ is invariant on each component of $M-F$ or $(2)$ there are exactly two component and $f$ interchanges them.

and after that he said:

In the case of $(2)$ the above argument shows that F cannot contain an open set and hence $dim F\leq (dim M) -1$ and since $F$ separates $M$ we have $dim F = (dim M) -1$. G. Bredon has shown that if $M$ is also orientable then any involution with an odd codimensional fixed point set must reverse the orientation; hence we obtain

Let $f$ be an orientation-preserving homemorphism of an orientable manifold $M$; then $f$ is invariant on each component of $M-F$.

Can you say me, what does mean the dim $F$ here? Is always $F$ is sub manifold with above condition? and how can we deduce that $dim F = n-1$?

For the first question, here is a sketch that $F$ is a submanifold under the assumption that the group $G$ acting on $M$ is finite:
Consider the map $M \to \prod_{g \in G} M, m \mapsto (gm)_{g \in G}$. This is smooth and should be a local homeomorphism, hence its regular. The diagonal $\{(m,\ldots,m) | m \in M\}$ of the product is a submanifold, hence its preimage is a submanifold of $M$ and the preimage is exactly the fixed point a set.
For the second question, we have that $M$ is connected but $M - F$, with $F$ being a submanifold is not connected. Intuitively it is clear that a submanifold dividing a manifold into connected components must have codimension 1 but I cannot think of a proof right now. Maybe one could work with path-connectedness?