Proof that the inverse image of a single element is a discrete spaceShow that in a discrete metric space, every subset is both open and closed.When is a local homeomorphism a covering map?Showing the fibre over a point in a covering map is a discrete space.Local homeomorphism and inverse imageHow do I show that a topological space is discrete if all its subsets are closed?Why is the Long Line not a covering space for the CircleFibers and they being discrete spaceLocal homeomorphisms which are not covering map?Alternative definition of covering spaces.Construction of a covering space as a fibre bundleProve that exist bijection between inverse image of covering spaceCovering spaces as fiber bundlesProof that a discrete space (with more than 1 element) is not connectedUniqueness of possibly non-simply connected universal covering spaces.Surjectivity of a (covering?) map
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Proof that the inverse image of a single element is a discrete space
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Proof that the inverse image of a single element is a discrete space
Show that in a discrete metric space, every subset is both open and closed.When is a local homeomorphism a covering map?Showing the fibre over a point in a covering map is a discrete space.Local homeomorphism and inverse imageHow do I show that a topological space is discrete if all its subsets are closed?Why is the Long Line not a covering space for the CircleFibers and they being discrete spaceLocal homeomorphisms which are not covering map?Alternative definition of covering spaces.Construction of a covering space as a fibre bundleProve that exist bijection between inverse image of covering spaceCovering spaces as fiber bundlesProof that a discrete space (with more than 1 element) is not connectedUniqueness of possibly non-simply connected universal covering spaces.Surjectivity of a (covering?) map
$begingroup$
Let $f: X rightarrow Y$ be a local homeomorphism. I want to prove that, for each $y in Y$, the fiber $f^-1(y)$ is a discrete set, or discrete space (Is there any difference between these two last terms?).
These are the posts I have read so far:
Local homeomorphism and inverse image
When is a local homeomorphism a covering map?
Showing the fibre over a point in a covering map is a discrete space.
How do I show that a topological space is discrete if all its subsets are closed?
Show that in a discrete metric space, every subset is both open and closed.
However, I have not been able to fully understand the proof. Some of the posts start the proof by mentioning that the fact that $f$ is a local homeomorphism implies that the fiber is finite; but I do not understand where does that come from, even after browsing MathSE and Wikipedia.
Other posts try instead to prove that the fiber is finite, and they do so by first stating that the fiber is a discrete space. All of this make it look like circular reasoning, which does not make sense to me.
If there is a concept I do not know, I am willing to visit places like Wikipedia or Subwiki.org ; however this time I have not been able to understand the proof even after reading many articles.
So, how can I prove this?
general-topology algebraic-topology metric-spaces topological-groups covering-spaces
asked 2 hours ago
evaristegdevaristegd
15110
$endgroup$
add a comment |
$begingroup$
Let $f: X rightarrow Y$ be a local homeomorphism. I want to prove that, for each $y in Y$, the fiber $f^-1(y)$ is a discrete set, or discrete space (Is there any difference between these two last terms?).
These are the posts I have read so far:
Local homeomorphism and inverse image
When is a local homeomorphism a covering map?
Showing the fibre over a point in a covering map is a discrete space.
How do I show that a topological space is discrete if all its subsets are closed?
Show that in a discrete metric space, every subset is both open and closed.
However, I have not been able to fully understand the proof. Some of the posts start the proof by mentioning that the fact that $f$ is a local homeomorphism implies that the fiber is finite; but I do not understand where does that come from, even after browsing MathSE and Wikipedia.
Other posts try instead to prove that the fiber is finite, and they do so by first stating that the fiber is a discrete space. All of this make it look like circular reasoning, which does not make sense to me.
If there is a concept I do not know, I am willing to visit places like Wikipedia or Subwiki.org ; however this time I have not been able to understand the proof even after reading many articles.
So, how can I prove this?
general-topology algebraic-topology metric-spaces topological-groups covering-spaces
asked 2 hours ago
evaristegdevaristegd
15110
$endgroup$
add a comment |
$begingroup$
Let $f: X rightarrow Y$ be a local homeomorphism. I want to prove that, for each $y in Y$, the fiber $f^-1(y)$ is a discrete set, or discrete space (Is there any difference between these two last terms?).
These are the posts I have read so far:
Local homeomorphism and inverse image
When is a local homeomorphism a covering map?
Showing the fibre over a point in a covering map is a discrete space.
How do I show that a topological space is discrete if all its subsets are closed?
Show that in a discrete metric space, every subset is both open and closed.
However, I have not been able to fully understand the proof. Some of the posts start the proof by mentioning that the fact that $f$ is a local homeomorphism implies that the fiber is finite; but I do not understand where does that come from, even after browsing MathSE and Wikipedia.
Other posts try instead to prove that the fiber is finite, and they do so by first stating that the fiber is a discrete space. All of this make it look like circular reasoning, which does not make sense to me.
If there is a concept I do not know, I am willing to visit places like Wikipedia or Subwiki.org ; however this time I have not been able to understand the proof even after reading many articles.
So, how can I prove this?
general-topology algebraic-topology metric-spaces topological-groups covering-spaces
asked 2 hours ago
evaristegdevaristegd
15110
$endgroup$
Let $f: X rightarrow Y$ be a local homeomorphism. I want to prove that, for each $y in Y$, the fiber $f^-1(y)$ is a discrete set, or discrete space (Is there any difference between these two last terms?).
These are the posts I have read so far:
Local homeomorphism and inverse image
When is a local homeomorphism a covering map?
Showing the fibre over a point in a covering map is a discrete space.
How do I show that a topological space is discrete if all its subsets are closed?
Show that in a discrete metric space, every subset is both open and closed.
However, I have not been able to fully understand the proof. Some of the posts start the proof by mentioning that the fact that $f$ is a local homeomorphism implies that the fiber is finite; but I do not understand where does that come from, even after browsing MathSE and Wikipedia.
Other posts try instead to prove that the fiber is finite, and they do so by first stating that the fiber is a discrete space. All of this make it look like circular reasoning, which does not make sense to me.
If there is a concept I do not know, I am willing to visit places like Wikipedia or Subwiki.org ; however this time I have not been able to understand the proof even after reading many articles.
So, how can I prove this?
general-topology algebraic-topology metric-spaces topological-groups covering-spaces
general-topology algebraic-topology metric-spaces topological-groups covering-spaces
asked 2 hours ago
evaristegdevaristegd
15110
asked 2 hours ago
evaristegdevaristegd
15110
asked 2 hours ago
evaristegdevaristegd
15110
asked 2 hours ago
evaristegdevaristegd
15110
asked 2 hours ago
evaristegdevaristegd
15110
15110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Fix $yin Y$ and $xin f^-1(y)$. Since $f$ is a local homeomorphism, there is a neighborhood $U$ of $x$ such that $f|U : Uto f(U)$ is a homeomorphism. If $zin Ucap f^-1(y)$, then $f(z) = y = f(x)$; since both $z, xin U$, injectivity of $f|U$ implies $z = x$. Therefore $Ucap f^-1(y) = x$. As $x$ was arbitrary, $f^-1(y)$ is discrete.
answered 2 hours ago
kobekobe
35.2k22248
$endgroup$
add a comment |
$begingroup$
Let $f:X to Y$ be a local homeomorphism.
Suppose $y in Y$ and let $x in F_y:=f^-1[y]$, the fibre of $y$.
Then $x$ has an open neighbourhood $U_x$ such that $f|_U_x: U_x to f[U_x]$ is a homeomorphism (by the definition of being a local homeomorphism).
In particular, $U_x cap F_y = x$ or else we have some $x' in U_x cap F_y$ which means $f(x')=f(x)=y$ while $x,x' in U_x$ contradicting the fact that $f|_U_x$ is injective (being a homeomorphism). So $U_x$ witnesses that $x$ is an isolated point of $F_y$, showing that $F_y$ is indeed discrete in the subspace topology.
Note that local injectivity is all we need.
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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$begingroup$
Fix $yin Y$ and $xin f^-1(y)$. Since $f$ is a local homeomorphism, there is a neighborhood $U$ of $x$ such that $f|U : Uto f(U)$ is a homeomorphism. If $zin Ucap f^-1(y)$, then $f(z) = y = f(x)$; since both $z, xin U$, injectivity of $f|U$ implies $z = x$. Therefore $Ucap f^-1(y) = x$. As $x$ was arbitrary, $f^-1(y)$ is discrete.
answered 2 hours ago
kobekobe
35.2k22248
$endgroup$
add a comment |
$begingroup$
Fix $yin Y$ and $xin f^-1(y)$. Since $f$ is a local homeomorphism, there is a neighborhood $U$ of $x$ such that $f|U : Uto f(U)$ is a homeomorphism. If $zin Ucap f^-1(y)$, then $f(z) = y = f(x)$; since both $z, xin U$, injectivity of $f|U$ implies $z = x$. Therefore $Ucap f^-1(y) = x$. As $x$ was arbitrary, $f^-1(y)$ is discrete.
answered 2 hours ago
kobekobe
35.2k22248
$endgroup$
add a comment |
$begingroup$
Fix $yin Y$ and $xin f^-1(y)$. Since $f$ is a local homeomorphism, there is a neighborhood $U$ of $x$ such that $f|U : Uto f(U)$ is a homeomorphism. If $zin Ucap f^-1(y)$, then $f(z) = y = f(x)$; since both $z, xin U$, injectivity of $f|U$ implies $z = x$. Therefore $Ucap f^-1(y) = x$. As $x$ was arbitrary, $f^-1(y)$ is discrete.
answered 2 hours ago
kobekobe
35.2k22248
$endgroup$
Fix $yin Y$ and $xin f^-1(y)$. Since $f$ is a local homeomorphism, there is a neighborhood $U$ of $x$ such that $f|U : Uto f(U)$ is a homeomorphism. If $zin Ucap f^-1(y)$, then $f(z) = y = f(x)$; since both $z, xin U$, injectivity of $f|U$ implies $z = x$. Therefore $Ucap f^-1(y) = x$. As $x$ was arbitrary, $f^-1(y)$ is discrete.
answered 2 hours ago
kobekobe
35.2k22248
answered 2 hours ago
kobekobe
35.2k22248
answered 2 hours ago
kobekobe
35.2k22248
answered 2 hours ago
kobekobe
35.2k22248
35.2k22248
add a comment |
add a comment |
$begingroup$
Let $f:X to Y$ be a local homeomorphism.
Suppose $y in Y$ and let $x in F_y:=f^-1[y]$, the fibre of $y$.
Then $x$ has an open neighbourhood $U_x$ such that $f|_U_x: U_x to f[U_x]$ is a homeomorphism (by the definition of being a local homeomorphism).
In particular, $U_x cap F_y = x$ or else we have some $x' in U_x cap F_y$ which means $f(x')=f(x)=y$ while $x,x' in U_x$ contradicting the fact that $f|_U_x$ is injective (being a homeomorphism). So $U_x$ witnesses that $x$ is an isolated point of $F_y$, showing that $F_y$ is indeed discrete in the subspace topology.
Note that local injectivity is all we need.
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
$endgroup$
add a comment |
$begingroup$
Let $f:X to Y$ be a local homeomorphism.
Suppose $y in Y$ and let $x in F_y:=f^-1[y]$, the fibre of $y$.
Then $x$ has an open neighbourhood $U_x$ such that $f|_U_x: U_x to f[U_x]$ is a homeomorphism (by the definition of being a local homeomorphism).
In particular, $U_x cap F_y = x$ or else we have some $x' in U_x cap F_y$ which means $f(x')=f(x)=y$ while $x,x' in U_x$ contradicting the fact that $f|_U_x$ is injective (being a homeomorphism). So $U_x$ witnesses that $x$ is an isolated point of $F_y$, showing that $F_y$ is indeed discrete in the subspace topology.
Note that local injectivity is all we need.
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
$endgroup$
add a comment |
$begingroup$
Let $f:X to Y$ be a local homeomorphism.
Suppose $y in Y$ and let $x in F_y:=f^-1[y]$, the fibre of $y$.
Then $x$ has an open neighbourhood $U_x$ such that $f|_U_x: U_x to f[U_x]$ is a homeomorphism (by the definition of being a local homeomorphism).
In particular, $U_x cap F_y = x$ or else we have some $x' in U_x cap F_y$ which means $f(x')=f(x)=y$ while $x,x' in U_x$ contradicting the fact that $f|_U_x$ is injective (being a homeomorphism). So $U_x$ witnesses that $x$ is an isolated point of $F_y$, showing that $F_y$ is indeed discrete in the subspace topology.
Note that local injectivity is all we need.
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
$endgroup$
Let $f:X to Y$ be a local homeomorphism.
Suppose $y in Y$ and let $x in F_y:=f^-1[y]$, the fibre of $y$.
Then $x$ has an open neighbourhood $U_x$ such that $f|_U_x: U_x to f[U_x]$ is a homeomorphism (by the definition of being a local homeomorphism).
In particular, $U_x cap F_y = x$ or else we have some $x' in U_x cap F_y$ which means $f(x')=f(x)=y$ while $x,x' in U_x$ contradicting the fact that $f|_U_x$ is injective (being a homeomorphism). So $U_x$ witnesses that $x$ is an isolated point of $F_y$, showing that $F_y$ is indeed discrete in the subspace topology.
Note that local injectivity is all we need.
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
answered 2 hours ago
Henno BrandsmaHenno Brandsma
119k351130
119k351130
add a comment |
add a comment |
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