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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 $dim F = n-1$?

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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?

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  • $\begingroup$ tahnk you so much $\endgroup$ – 123... Jun 10 '13 at 6:58
  • $\begingroup$ Excuse me, I have not understand your argument yet. in your argument you said the above map is local homeomorphism, it can not be happen, because the co domain has greater dim than domain. furthermore, what is the meaning of regular for homeomorphism between topological manifold? $\endgroup$ – 123... Jun 14 '13 at 7:48
  • $\begingroup$ Ok you are right, my idea is just plainly wrong. I missed the point that we are talking about topological manifolds only, so the concept of regularity is not available. Do you have any conditions on your manifold or the group action or could you say which paper you are talking about? $\endgroup$ – Alexander Körschgen Jun 15 '13 at 13:28
  • $\begingroup$ Thank you so much for your taking time on my question. $\endgroup$ – 123... Jun 16 '13 at 16:56
  • $\begingroup$ Thank you so much for taking your time to answer my question. I think we must solve this problem for involution between topological manifold. I know the solution for the smooth (involution) case. this problem is in the following paper "M. Brown and J.M. Kister, Invariance of complementary domains of a fixed point set, Proc. Amer. Math. Soc. 91, No 3, (1984), 503-504." $\endgroup$ – 123... Jun 16 '13 at 17:14

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