Involution and Covering space The Next CEO of Stack OverflowExamples of involutions on $mathbbR$The covering space of connected spacehelix and covering space of the unit circleRigorous Covering Space ConstructionHow to find n-sheeted covering space of a topological space explicitly(If exists)?Finite fundamental group and covering spacesCovering space induced by properly discontinuous group actionConstructing a universal covering spaceIf $X$ is path-connected and has a finite fundamental group then every covering map from $X$ to $X$ is an homeomorphism.Show that a continuous map from covering space is a homeomorphism if fundamental group is finite.What is a normal covering geometrically?
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Involution and Covering space
The Next CEO of Stack OverflowExamples of involutions on $mathbbR$The covering space of connected spacehelix and covering space of the unit circleRigorous Covering Space ConstructionHow to find n-sheeted covering space of a topological space explicitly(If exists)?Finite fundamental group and covering spacesCovering space induced by properly discontinuous group actionConstructing a universal covering spaceIf $X$ is path-connected and has a finite fundamental group then every covering map from $X$ to $X$ is an homeomorphism.Show that a continuous map from covering space is a homeomorphism if fundamental group is finite.What is a normal covering geometrically?
$begingroup$
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?
covering-spaces involutions
asked Oct 3 '15 at 7:20
123...123...
426213
$endgroup$
add a comment |
$begingroup$
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?
covering-spaces involutions
asked Oct 3 '15 at 7:20
123...123...
426213
$endgroup$
1
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51
add a comment |
$begingroup$
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?
covering-spaces involutions
asked Oct 3 '15 at 7:20
123...123...
426213
$endgroup$
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?
covering-spaces involutions
covering-spaces involutions
asked Oct 3 '15 at 7:20
123...123...
426213
asked Oct 3 '15 at 7:20
123...123...
426213
asked Oct 3 '15 at 7:20
123...123...
426213
asked Oct 3 '15 at 7:20
123...123...
426213
asked Oct 3 '15 at 7:20
123...123...
426213
426213
1
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51
add a comment |
1
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51
1
1
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $mathbbR$ 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 $mathbbR$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
add a comment |
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$begingroup$
A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $mathbbR$ 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 $mathbbR$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
add a comment |
$begingroup$
A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $mathbbR$ 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 $mathbbR$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
add a comment |
$begingroup$
A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $mathbbR$ 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 $mathbbR$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
A simply connected topological space is its own universal cover, so your question reduces to the following:
Does $mathbbR$ 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 $mathbbR$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
edited yesterday
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
answered Aug 9 '17 at 16:14
Michael AlbaneseMichael Albanese
64.3k1599313
64.3k1599313
add a comment |
add a comment |
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1
$begingroup$
please define a free involution.
$endgroup$
– R_D
Oct 3 '15 at 8:09
$begingroup$
A free involution $nu : Xto X $ on a topological space $X$ is a fixed point free homeomorphism such that $nu onu = Id_X$
$endgroup$
– 123...
Oct 3 '15 at 9:51