Is there a connected topological space such that admits a free involution, trivial fundamental group and furthermore has the set of real number as it's covering space?
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1please define a free involution. – R_D Oct 03 '15 at 08:09
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A free involution $\nu : X\to X $ on a topological space $X$ is a fixed point free homeomorphism such that $\nu o\nu = Id_{X}$ – 123... Oct 03 '15 at 09:51
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A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $\mathbb{R}$ admit a free involution $f$?
I'm going to assume you want the involution $f$ to be continuous. If you don't require continuity, then there are many examples
If $f$ has no fixed points, it is injective and is therefore either increasing or decreasing. If it were increasing, we'd have $x < f(x) < f(f(x)) = x$ which is a contradiction. If it were instead decreasing, we would find that $x = f(f(x)) < f(x) < x$ which is again a contradiction.
Therefore, every continuous involution of $\mathbb{R}$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
Michael Albanese
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