Fibers Under a Covering Map are Discrete Subspaces of the Domain
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In Munkres' topology book, the following claim is made:
If $p : E to B$ is a covering map, then for every $b in B$, $p^{-1}(b)$ is a discrete subspace of $E$.
Here's my attempt at a proof:
Given $b in B$, there's an open set $U ni b$ and disjoint open sets ${V_i}_{i in I}$ in $X$ such that $p^{-1}(U) = bigcup_{i in I} V_i$ and such that $p big|_{V_i} : V_i to U$ is a homeomorphism. Hence, if $x in p^{-1}(b) subseteq p^{-1}(U)$, then $x in V_i$ for some $i in I$. If $y in p^{-1}(b) cap V_i$, then $p(y) = b$ and $y in V_i$. But $p big|_{V_i}$ is, in particular, injective so $p big|_{V_i}(x) = b = p big|_{V_i}(y)$ implies $x=y$. This shows $p^{-1}(b) cap V_i = {x}$, proving that ${x}$ is open in $p^{-1}(b)$.
Does this seem right?
general-topology algebraic-topology covering-spaces
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
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add a comment |
$begingroup$
In Munkres' topology book, the following claim is made:
If $p : E to B$ is a covering map, then for every $b in B$, $p^{-1}(b)$ is a discrete subspace of $E$.
Here's my attempt at a proof:
Given $b in B$, there's an open set $U ni b$ and disjoint open sets ${V_i}_{i in I}$ in $X$ such that $p^{-1}(U) = bigcup_{i in I} V_i$ and such that $p big|_{V_i} : V_i to U$ is a homeomorphism. Hence, if $x in p^{-1}(b) subseteq p^{-1}(U)$, then $x in V_i$ for some $i in I$. If $y in p^{-1}(b) cap V_i$, then $p(y) = b$ and $y in V_i$. But $p big|_{V_i}$ is, in particular, injective so $p big|_{V_i}(x) = b = p big|_{V_i}(y)$ implies $x=y$. This shows $p^{-1}(b) cap V_i = {x}$, proving that ${x}$ is open in $p^{-1}(b)$.
Does this seem right?
general-topology algebraic-topology covering-spaces
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
$endgroup$
$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59
add a comment |
$begingroup$
In Munkres' topology book, the following claim is made:
If $p : E to B$ is a covering map, then for every $b in B$, $p^{-1}(b)$ is a discrete subspace of $E$.
Here's my attempt at a proof:
Given $b in B$, there's an open set $U ni b$ and disjoint open sets ${V_i}_{i in I}$ in $X$ such that $p^{-1}(U) = bigcup_{i in I} V_i$ and such that $p big|_{V_i} : V_i to U$ is a homeomorphism. Hence, if $x in p^{-1}(b) subseteq p^{-1}(U)$, then $x in V_i$ for some $i in I$. If $y in p^{-1}(b) cap V_i$, then $p(y) = b$ and $y in V_i$. But $p big|_{V_i}$ is, in particular, injective so $p big|_{V_i}(x) = b = p big|_{V_i}(y)$ implies $x=y$. This shows $p^{-1}(b) cap V_i = {x}$, proving that ${x}$ is open in $p^{-1}(b)$.
Does this seem right?
general-topology algebraic-topology covering-spaces
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
$endgroup$
In Munkres' topology book, the following claim is made:
If $p : E to B$ is a covering map, then for every $b in B$, $p^{-1}(b)$ is a discrete subspace of $E$.
Here's my attempt at a proof:
Given $b in B$, there's an open set $U ni b$ and disjoint open sets ${V_i}_{i in I}$ in $X$ such that $p^{-1}(U) = bigcup_{i in I} V_i$ and such that $p big|_{V_i} : V_i to U$ is a homeomorphism. Hence, if $x in p^{-1}(b) subseteq p^{-1}(U)$, then $x in V_i$ for some $i in I$. If $y in p^{-1}(b) cap V_i$, then $p(y) = b$ and $y in V_i$. But $p big|_{V_i}$ is, in particular, injective so $p big|_{V_i}(x) = b = p big|_{V_i}(y)$ implies $x=y$. This shows $p^{-1}(b) cap V_i = {x}$, proving that ${x}$ is open in $p^{-1}(b)$.
Does this seem right?
general-topology algebraic-topology covering-spaces
general-topology algebraic-topology covering-spaces
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
asked Dec 11 '18 at 22:42
user193319user193319
2,4102925
2,4102925
$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59
add a comment |
$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59
$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59
add a comment |
1 Answer
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I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
$endgroup$
I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
answered Dec 12 '18 at 1:10
zoidbergzoidberg
1,080113
1,080113
add a comment |
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$begingroup$
It seems fine for me, but let the experts talk.
$endgroup$
– Dog_69
Dec 11 '18 at 22:48
$begingroup$
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate by projection.
$endgroup$
– zoidberg
Dec 11 '18 at 23:59