Space with semi-locally simply connected open subsets
$begingroup$
A topological space $X$ is semi-locally simply connected if, for any $xin X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$Pi_1(U)rightarrowPi_1(X)$$ induced by the inclusion $Usubseteq X$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.
My question is: is it true that if a locally path connected space $X$ is such that, for any open subset $Usubseteq X$, $U$ is semi-locally simply connected, then $X$ must be locally simply connected?
general-topology algebraic-topology homotopy-theory
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
$endgroup$
|
show 9 more comments
$begingroup$
A topological space $X$ is semi-locally simply connected if, for any $xin X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$Pi_1(U)rightarrowPi_1(X)$$ induced by the inclusion $Usubseteq X$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.
My question is: is it true that if a locally path connected space $X$ is such that, for any open subset $Usubseteq X$, $U$ is semi-locally simply connected, then $X$ must be locally simply connected?
general-topology algebraic-topology homotopy-theory
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
$endgroup$
$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
1
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
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You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
1
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
1
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17
|
show 9 more comments
$begingroup$
A topological space $X$ is semi-locally simply connected if, for any $xin X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$Pi_1(U)rightarrowPi_1(X)$$ induced by the inclusion $Usubseteq X$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.
My question is: is it true that if a locally path connected space $X$ is such that, for any open subset $Usubseteq X$, $U$ is semi-locally simply connected, then $X$ must be locally simply connected?
general-topology algebraic-topology homotopy-theory
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
$endgroup$
A topological space $X$ is semi-locally simply connected if, for any $xin X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$Pi_1(U)rightarrowPi_1(X)$$ induced by the inclusion $Usubseteq X$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.
My question is: is it true that if a locally path connected space $X$ is such that, for any open subset $Usubseteq X$, $U$ is semi-locally simply connected, then $X$ must be locally simply connected?
general-topology algebraic-topology homotopy-theory
general-topology algebraic-topology homotopy-theory
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
edited Dec 19 '18 at 20:52
mfox
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
asked Oct 21 '18 at 16:01
mfoxmfox
3101314
3101314
$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
1
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
$begingroup$
You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
1
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
1
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17
|
show 9 more comments
$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
1
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
$begingroup$
You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
1
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
1
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17
$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
1
1
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
$begingroup$
You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
$begingroup$
You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
1
1
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
1
1
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17
|
show 9 more comments
1 Answer
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$begingroup$
Consider the subspace of $mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $xin{0}cup{1/n:ninmathbb{Z},n>0}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $pi_0$ and $pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
$endgroup$
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
add a comment |
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$begingroup$
Consider the subspace of $mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $xin{0}cup{1/n:ninmathbb{Z},n>0}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $pi_0$ and $pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
$endgroup$
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
add a comment |
$begingroup$
Consider the subspace of $mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $xin{0}cup{1/n:ninmathbb{Z},n>0}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $pi_0$ and $pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
$endgroup$
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
add a comment |
$begingroup$
Consider the subspace of $mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $xin{0}cup{1/n:ninmathbb{Z},n>0}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $pi_0$ and $pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
$endgroup$
Consider the subspace of $mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $xin{0}cup{1/n:ninmathbb{Z},n>0}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $pi_0$ and $pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
edited Oct 30 '18 at 14:32
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
answered Oct 23 '18 at 2:31
Kyle MillerKyle Miller
9,088929
9,088929
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
add a comment |
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
1
1
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
$CX$ is semilocally simply connected as you say, but it's not true that each one of its open subset is semilocally simply connected ($Xtimes [0, 1)$ for example is not semilocally simply connected), so $CX$ is not a valid counterexample.
$endgroup$
– mfox
Oct 23 '18 at 8:35
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
$begingroup$
@mfox I misunderstood your question because of the re-use of the letter $U$ (I should have read it more carefully!) I've replaced the answer, though I'd much rather have a counterexample that doesn't feel like it's getting by on a technicality.
$endgroup$
– Kyle Miller
Oct 30 '18 at 14:38
add a comment |
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$begingroup$
Sorry, I've misread.
$endgroup$
– Saucy O'Path
Oct 21 '18 at 16:29
1
$begingroup$
Your equivalent reformulation of "any loop in $U$ is homotopic to a constant one in $X$" is only valid if $U$ is pathconnected : you should say that the image of this functor has at most one arrow between any two points
$endgroup$
– Max
Oct 23 '18 at 9:32
$begingroup$
You're right, in my mind I was assuming without a reason that $X$ is locally path connected. Thank you, I will make an edit.
$endgroup$
– mfox
Oct 23 '18 at 9:42
1
$begingroup$
@user126154 The two-point discrete topology is not simply connected (since it's not path connected), yet still every open subset is semilocally simply connected.
$endgroup$
– Kyle Miller
Dec 20 '18 at 18:52
1
$begingroup$
@user126154 Every open set of a manifold is semilocally simply connected (since manifolds are locally simply connected), but many manifolds aren't simply connected.
$endgroup$
– Kyle Miller
Dec 21 '18 at 20:17