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3

$begingroup$

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 metric-spaces covering-spaces

share|cite|improve this question

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

$endgroup$

add a comment | 

3

$begingroup$

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 metric-spaces covering-spaces

share|cite|improve this question

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

$endgroup$

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 metric-spaces covering-spaces

share|cite|improve this question

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

$endgroup$

share|cite|improve this question

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

share|cite|improve this question

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

edited May 12 at 6:14

Henno Brandsma

120k351130

edited May 12 at 6:14

Henno Brandsma

120k351130

asked May 11 at 20:17

evaristegdevaristegd

16511

asked May 11 at 20:17

evaristegdevaristegd

16511

2 Answers
2

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

kobekobe

35.3k22248

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

kobekobe

35.3k22248

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

kobekobe

35.3k22248

$endgroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

kobekobe

35.3k22248

$endgroup$

share|cite|improve this answer

answered May 11 at 20:30

kobekobe

35.3k22248

answered May 11 at 20:30

kobekobe

35.3k22248

answered May 11 at 20:30

kobekobe

35.3k22248

4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

$endgroup$

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.

share|cite|improve this answer

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

$endgroup$

share|cite|improve this answer

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130

answered May 11 at 20:30

Henno BrandsmaHenno Brandsma

120k351130