the fundamental group of $X$ is the symmetric group $S_3$, then whether it has a universal cover?
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Question: Suppose that $X$ is a path-connected space with $pi_1(X)=S_3$, which is the 3-symmetric group. I just wonder that whether $X$ has a universal cover.
Try: Based on Hatcher, $X$ has a universal cover iff $X$ is path-connected, locally path-connected, and semilocally simply-connected. However, to prove that $X$ meets the latter two conditions is not easy.
I know that $X$ with $pi_1(X)=S_3$ can be realized by a CW complex and any CW complex meets these three conditions so that has a universal cover. But this is not the way to show that any such $X$ has a universal cover.
algebraic-topology covering-spaces fundamental-groups path-connected
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
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add a comment |
$begingroup$
Question: Suppose that $X$ is a path-connected space with $pi_1(X)=S_3$, which is the 3-symmetric group. I just wonder that whether $X$ has a universal cover.
Try: Based on Hatcher, $X$ has a universal cover iff $X$ is path-connected, locally path-connected, and semilocally simply-connected. However, to prove that $X$ meets the latter two conditions is not easy.
I know that $X$ with $pi_1(X)=S_3$ can be realized by a CW complex and any CW complex meets these three conditions so that has a universal cover. But this is not the way to show that any such $X$ has a universal cover.
algebraic-topology covering-spaces fundamental-groups path-connected
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
$endgroup$
add a comment |
$begingroup$
Question: Suppose that $X$ is a path-connected space with $pi_1(X)=S_3$, which is the 3-symmetric group. I just wonder that whether $X$ has a universal cover.
Try: Based on Hatcher, $X$ has a universal cover iff $X$ is path-connected, locally path-connected, and semilocally simply-connected. However, to prove that $X$ meets the latter two conditions is not easy.
I know that $X$ with $pi_1(X)=S_3$ can be realized by a CW complex and any CW complex meets these three conditions so that has a universal cover. But this is not the way to show that any such $X$ has a universal cover.
algebraic-topology covering-spaces fundamental-groups path-connected
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
$endgroup$
Question: Suppose that $X$ is a path-connected space with $pi_1(X)=S_3$, which is the 3-symmetric group. I just wonder that whether $X$ has a universal cover.
Try: Based on Hatcher, $X$ has a universal cover iff $X$ is path-connected, locally path-connected, and semilocally simply-connected. However, to prove that $X$ meets the latter two conditions is not easy.
I know that $X$ with $pi_1(X)=S_3$ can be realized by a CW complex and any CW complex meets these three conditions so that has a universal cover. But this is not the way to show that any such $X$ has a universal cover.
algebraic-topology covering-spaces fundamental-groups path-connected
algebraic-topology covering-spaces fundamental-groups path-connected
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
asked Nov 28 '18 at 16:38
QUAN CHENQUAN CHEN
1598
1598
add a comment |
add a comment |
1 Answer
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$begingroup$
Knowledge of $pi_1(X)$ and nothing else will not tell you that $X$ is path-connected, locally so, or semilocally simply-connected. Since Hatcher's criteria is an "if and only if" your question has no definite answer as written. The closest you may be able to get is via a CW-approximation which would be correct up to (weak) homotopy, but would not say anything about $X$ itself.
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
$endgroup$
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Knowledge of $pi_1(X)$ and nothing else will not tell you that $X$ is path-connected, locally so, or semilocally simply-connected. Since Hatcher's criteria is an "if and only if" your question has no definite answer as written. The closest you may be able to get is via a CW-approximation which would be correct up to (weak) homotopy, but would not say anything about $X$ itself.
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
$endgroup$
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
add a comment |
$begingroup$
Knowledge of $pi_1(X)$ and nothing else will not tell you that $X$ is path-connected, locally so, or semilocally simply-connected. Since Hatcher's criteria is an "if and only if" your question has no definite answer as written. The closest you may be able to get is via a CW-approximation which would be correct up to (weak) homotopy, but would not say anything about $X$ itself.
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
$endgroup$
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
add a comment |
$begingroup$
Knowledge of $pi_1(X)$ and nothing else will not tell you that $X$ is path-connected, locally so, or semilocally simply-connected. Since Hatcher's criteria is an "if and only if" your question has no definite answer as written. The closest you may be able to get is via a CW-approximation which would be correct up to (weak) homotopy, but would not say anything about $X$ itself.
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
$endgroup$
Knowledge of $pi_1(X)$ and nothing else will not tell you that $X$ is path-connected, locally so, or semilocally simply-connected. Since Hatcher's criteria is an "if and only if" your question has no definite answer as written. The closest you may be able to get is via a CW-approximation which would be correct up to (weak) homotopy, but would not say anything about $X$ itself.
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
answered Nov 28 '18 at 16:42
RandallRandall
9,71111230
9,71111230
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
add a comment |
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
$begingroup$
we have supposed that $X$ is path connected.
$endgroup$
– QUAN CHEN
Nov 28 '18 at 16:58
1
1
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
$begingroup$
Right, but then you need to get locally p-c and semilocally s-c, and all of these properties are independent.
$endgroup$
– Randall
Nov 28 '18 at 17:53
add a comment |
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