Hatcher Universal Covering Space Construction - Basis
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Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:
I don't understand why it follows that $mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.
general-topology algebraic-topology
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
$endgroup$
add a comment |
$begingroup$
Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:
I don't understand why it follows that $mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.
general-topology algebraic-topology
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
$endgroup$
$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
$endgroup$
– Randall
Dec 9 '18 at 20:41
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I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
$begingroup$
That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57
add a comment |
$begingroup$
Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:
I don't understand why it follows that $mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.
general-topology algebraic-topology
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
$endgroup$
Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:
I don't understand why it follows that $mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.
general-topology algebraic-topology
general-topology algebraic-topology
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
asked Dec 9 '18 at 20:36
Frederic ChopinFrederic Chopin
328111
328111
$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
$endgroup$
– Randall
Dec 9 '18 at 20:41
$begingroup$
I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
$begingroup$
That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57
add a comment |
$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
$endgroup$
– Randall
Dec 9 '18 at 20:41
$begingroup$
I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
$begingroup$
That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57
$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
$endgroup$
– Randall
Dec 9 '18 at 20:41
$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
$endgroup$
– Randall
Dec 9 '18 at 20:41
$begingroup$
I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
$begingroup$
I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
$begingroup$
That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57
$begingroup$
That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A basis for a topology $mathcal{T}$ on a space $X$ is a subset $mathcal{B} subset mathcal{T}$ such that for each $U in mathcal{T}$ and each $x in U$ there exists $B in mathcal{B}$ such that $x in B subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $mathcal{U} subset mathcal{P}$ of all $U in mathcal{P}$ such that $pi_1(U) to pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
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– Paul Frost
Dec 10 '18 at 11:50
add a comment |
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Let me iron all of this out.
Let $U$ be an open set of $X$ and $x in U$. Since $X$ is locally path connected, $X$ has a basis $mathcal{B}$ of path connected sets. Hence, we can find a $B_1 in mathcal{B}$ such that $x in B_1 subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $pi_1(V,x) rightarrow pi_1(X,x)$ is trivial. Hence, we can find a $B_2 in mathcal{B}$ such that $x in B_2 subset V$ and furthermore we can find a $B_3 in mathcal{B}$ such that $x in B_3 subset B_1 cap B_2 subset V$. Then the composition of induced inclusions $pi_1(B_3,x) rightarrow pi_1(V,x) rightarrow pi_1(X,x)$ is trivial and $B_3 subset U$. Thus, what we have shown is that the subset of $mathcal{B}$ consisting of elements $W$ such that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial is also a basis for $X$.
(Remember that since $W$ is path connected and the induced inclusion $pi_1(W,x) rightarrow pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial.)
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
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$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
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– Frederic Chopin
Dec 10 '18 at 2:46
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A basis for a topology $mathcal{T}$ on a space $X$ is a subset $mathcal{B} subset mathcal{T}$ such that for each $U in mathcal{T}$ and each $x in U$ there exists $B in mathcal{B}$ such that $x in B subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $mathcal{U} subset mathcal{P}$ of all $U in mathcal{P}$ such that $pi_1(U) to pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
add a comment |
$begingroup$
A basis for a topology $mathcal{T}$ on a space $X$ is a subset $mathcal{B} subset mathcal{T}$ such that for each $U in mathcal{T}$ and each $x in U$ there exists $B in mathcal{B}$ such that $x in B subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $mathcal{U} subset mathcal{P}$ of all $U in mathcal{P}$ such that $pi_1(U) to pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
$endgroup$
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
add a comment |
$begingroup$
A basis for a topology $mathcal{T}$ on a space $X$ is a subset $mathcal{B} subset mathcal{T}$ such that for each $U in mathcal{T}$ and each $x in U$ there exists $B in mathcal{B}$ such that $x in B subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $mathcal{U} subset mathcal{P}$ of all $U in mathcal{P}$ such that $pi_1(U) to pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
$endgroup$
A basis for a topology $mathcal{T}$ on a space $X$ is a subset $mathcal{B} subset mathcal{T}$ such that for each $U in mathcal{T}$ and each $x in U$ there exists $B in mathcal{B}$ such that $x in B subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $mathcal{U} subset mathcal{P}$ of all $U in mathcal{P}$ such that $pi_1(U) to pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
answered Dec 9 '18 at 22:43
Paul FrostPaul Frost
10.3k3933
10.3k3933
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
add a comment |
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x in X$, there is a neighbourhood of $x$ which is path connected.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:45
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x in V subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x in U$, there is a basis element $B$ such that $x in B subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions.
$endgroup$
– Frederic Chopin
Dec 9 '18 at 23:48
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 0:16
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
$begingroup$
If you require that each $x in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:36
1
1
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
$begingroup$
Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $forall x in X$ $ forall$ neighborhoods $U$ of $x$ $dots$ Now consider any open $U$ and any $x in U$. Then $x in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $forall$ open $U$ $forall x in U dots$ Now consider any $x in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x in U$, so (2) applies.
$endgroup$
– Paul Frost
Dec 10 '18 at 11:50
add a comment |
$begingroup$
Let me iron all of this out.
Let $U$ be an open set of $X$ and $x in U$. Since $X$ is locally path connected, $X$ has a basis $mathcal{B}$ of path connected sets. Hence, we can find a $B_1 in mathcal{B}$ such that $x in B_1 subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $pi_1(V,x) rightarrow pi_1(X,x)$ is trivial. Hence, we can find a $B_2 in mathcal{B}$ such that $x in B_2 subset V$ and furthermore we can find a $B_3 in mathcal{B}$ such that $x in B_3 subset B_1 cap B_2 subset V$. Then the composition of induced inclusions $pi_1(B_3,x) rightarrow pi_1(V,x) rightarrow pi_1(X,x)$ is trivial and $B_3 subset U$. Thus, what we have shown is that the subset of $mathcal{B}$ consisting of elements $W$ such that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial is also a basis for $X$.
(Remember that since $W$ is path connected and the induced inclusion $pi_1(W,x) rightarrow pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial.)
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
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$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
add a comment |
$begingroup$
Let me iron all of this out.
Let $U$ be an open set of $X$ and $x in U$. Since $X$ is locally path connected, $X$ has a basis $mathcal{B}$ of path connected sets. Hence, we can find a $B_1 in mathcal{B}$ such that $x in B_1 subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $pi_1(V,x) rightarrow pi_1(X,x)$ is trivial. Hence, we can find a $B_2 in mathcal{B}$ such that $x in B_2 subset V$ and furthermore we can find a $B_3 in mathcal{B}$ such that $x in B_3 subset B_1 cap B_2 subset V$. Then the composition of induced inclusions $pi_1(B_3,x) rightarrow pi_1(V,x) rightarrow pi_1(X,x)$ is trivial and $B_3 subset U$. Thus, what we have shown is that the subset of $mathcal{B}$ consisting of elements $W$ such that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial is also a basis for $X$.
(Remember that since $W$ is path connected and the induced inclusion $pi_1(W,x) rightarrow pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial.)
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
$endgroup$
$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
add a comment |
$begingroup$
Let me iron all of this out.
Let $U$ be an open set of $X$ and $x in U$. Since $X$ is locally path connected, $X$ has a basis $mathcal{B}$ of path connected sets. Hence, we can find a $B_1 in mathcal{B}$ such that $x in B_1 subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $pi_1(V,x) rightarrow pi_1(X,x)$ is trivial. Hence, we can find a $B_2 in mathcal{B}$ such that $x in B_2 subset V$ and furthermore we can find a $B_3 in mathcal{B}$ such that $x in B_3 subset B_1 cap B_2 subset V$. Then the composition of induced inclusions $pi_1(B_3,x) rightarrow pi_1(V,x) rightarrow pi_1(X,x)$ is trivial and $B_3 subset U$. Thus, what we have shown is that the subset of $mathcal{B}$ consisting of elements $W$ such that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial is also a basis for $X$.
(Remember that since $W$ is path connected and the induced inclusion $pi_1(W,x) rightarrow pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial.)
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
$endgroup$
Let me iron all of this out.
Let $U$ be an open set of $X$ and $x in U$. Since $X$ is locally path connected, $X$ has a basis $mathcal{B}$ of path connected sets. Hence, we can find a $B_1 in mathcal{B}$ such that $x in B_1 subset U$. Since $X$ is semilocally simply connected, there is a an open set $V$ containing $x$ such that the induced inclusion $pi_1(V,x) rightarrow pi_1(X,x)$ is trivial. Hence, we can find a $B_2 in mathcal{B}$ such that $x in B_2 subset V$ and furthermore we can find a $B_3 in mathcal{B}$ such that $x in B_3 subset B_1 cap B_2 subset V$. Then the composition of induced inclusions $pi_1(B_3,x) rightarrow pi_1(V,x) rightarrow pi_1(X,x)$ is trivial and $B_3 subset U$. Thus, what we have shown is that the subset of $mathcal{B}$ consisting of elements $W$ such that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial is also a basis for $X$.
(Remember that since $W$ is path connected and the induced inclusion $pi_1(W,x) rightarrow pi_1(X,x)$ is trivial for the choice of basepoint $x$ above, the induced inclusion is trivial for all other basepoints in $W$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $pi_1(W) rightarrow pi_1(X)$ is trivial.)
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
answered Dec 10 '18 at 1:41
Frederic ChopinFrederic Chopin
328111
328111
$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
add a comment |
$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
$begingroup$
I have added a follow-up clarification here: math.stackexchange.com/questions/3033345/…
$endgroup$
– Frederic Chopin
Dec 10 '18 at 2:46
add a comment |
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$begingroup$
Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book.
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– Randall
Dec 9 '18 at 20:41
$begingroup$
I haven't seen this before. The condition I'm familiar with is that given any open set $U subset X$, for all $x in U$, there is a basis element $V$ such that $x in V subset U$. Is that fact related to this condition?
$endgroup$
– Frederic Chopin
Dec 9 '18 at 20:43
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That is what’s underlined in red.
$endgroup$
– Randall
Dec 9 '18 at 20:57