help me in trace of following proposition In a paper an author proved the following proposition
Please help me in  trace proof of 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?
