Showing the image of a surjective map from a regular topological space is also regular
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Problem : Suppose that $X$ is a regular topological space, $Y$ a topological space. If there exists a continuous surjective map $f$ from $X rightarrow Y$ that is closed, where $f^{-1}(y)$ is compact for every $y in Y$, then I want to show that $Y$ is regular as well.
Thoughts:
First, I was thinking to use the characterization of a regular space that :
a topological space $X$ is regular iff for every $x in X$, and every neighbourhood $U$ of $x$, there exists a neighbourhood $V$ such that $x in V subset cl(V) subset X$.
So for $Y$ in this case take a point $y in Y$, then $f^{-1}(y)$ is non-empty by surjectivity. Take $x$ in $f^{-1}(y)$, and then by the regularity of $X$ there are neighbourhoods $U,V$ of $x$ such that $x in V in cl(V) in U$ where $f(cl(V))$ is a closed neighbourhood of $y$, say $C$. This is about as far as I get and I am not sure how to proceed.
general-topology
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
$endgroup$
add a comment |
$begingroup$
Problem : Suppose that $X$ is a regular topological space, $Y$ a topological space. If there exists a continuous surjective map $f$ from $X rightarrow Y$ that is closed, where $f^{-1}(y)$ is compact for every $y in Y$, then I want to show that $Y$ is regular as well.
Thoughts:
First, I was thinking to use the characterization of a regular space that :
a topological space $X$ is regular iff for every $x in X$, and every neighbourhood $U$ of $x$, there exists a neighbourhood $V$ such that $x in V subset cl(V) subset X$.
So for $Y$ in this case take a point $y in Y$, then $f^{-1}(y)$ is non-empty by surjectivity. Take $x$ in $f^{-1}(y)$, and then by the regularity of $X$ there are neighbourhoods $U,V$ of $x$ such that $x in V in cl(V) in U$ where $f(cl(V))$ is a closed neighbourhood of $y$, say $C$. This is about as far as I get and I am not sure how to proceed.
general-topology
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
$endgroup$
add a comment |
$begingroup$
Problem : Suppose that $X$ is a regular topological space, $Y$ a topological space. If there exists a continuous surjective map $f$ from $X rightarrow Y$ that is closed, where $f^{-1}(y)$ is compact for every $y in Y$, then I want to show that $Y$ is regular as well.
Thoughts:
First, I was thinking to use the characterization of a regular space that :
a topological space $X$ is regular iff for every $x in X$, and every neighbourhood $U$ of $x$, there exists a neighbourhood $V$ such that $x in V subset cl(V) subset X$.
So for $Y$ in this case take a point $y in Y$, then $f^{-1}(y)$ is non-empty by surjectivity. Take $x$ in $f^{-1}(y)$, and then by the regularity of $X$ there are neighbourhoods $U,V$ of $x$ such that $x in V in cl(V) in U$ where $f(cl(V))$ is a closed neighbourhood of $y$, say $C$. This is about as far as I get and I am not sure how to proceed.
general-topology
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
$endgroup$
Problem : Suppose that $X$ is a regular topological space, $Y$ a topological space. If there exists a continuous surjective map $f$ from $X rightarrow Y$ that is closed, where $f^{-1}(y)$ is compact for every $y in Y$, then I want to show that $Y$ is regular as well.
Thoughts:
First, I was thinking to use the characterization of a regular space that :
a topological space $X$ is regular iff for every $x in X$, and every neighbourhood $U$ of $x$, there exists a neighbourhood $V$ such that $x in V subset cl(V) subset X$.
So for $Y$ in this case take a point $y in Y$, then $f^{-1}(y)$ is non-empty by surjectivity. Take $x$ in $f^{-1}(y)$, and then by the regularity of $X$ there are neighbourhoods $U,V$ of $x$ such that $x in V in cl(V) in U$ where $f(cl(V))$ is a closed neighbourhood of $y$, say $C$. This is about as far as I get and I am not sure how to proceed.
general-topology
general-topology
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
asked Nov 28 '18 at 17:31
IntegrateThisIntegrateThis
1,7861718
1,7861718
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X to Y$ is a closed continuous map.
- Then given $y in Y$ and an open $U subseteq X$ with $f^{-1} ( y ) subseteq U$ there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U$.
- Then $overline{ f(A) } = f ( overline{A} )$ for all $A subseteq X$.
(In fact, both of the above are equivalent characterizations of closed mappings.)
Pick $y in Y$ and an open neighbourhood $V subseteq Y$ of $y$. Then $K = f^{-1} ( y ) subseteq X$ is compact and $U = f^{-1} ( V ) subseteq X$ is open and $K subseteq U$.
By regularity of $X$ for each $x in K$ there is an open $U_x subseteq X$ with $x in U_x subseteq overline{U_x} subseteq U$. As ${ U_x : x in K }$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , ldots , x_n in K$ such that $K subseteq U_{x_1} cup cdots cup U_{x_n}$. Note that $overline{U_{x_1} cup cdots cup U_{x_n}} = overline{U_{x_1}} cup cdots cup overline{U_{x_n}} subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U_{x_1} cup cdots cup U_{x_n}$.
Using Fact(2) above (and the surjectivity of $f$) it follows that $$overline{W} = overline{f(f^{-1} (W)} = f ( overline{f^{-1}(W) } ) subseteq f ( overline{ U_{x_1} cup cdots cup U_{x_n} } ) subseteq f ( U ) = V.$$
That is, we have found an open $W subseteq Y$ such that $y in W subseteq overline{W} subseteq V$.
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
$endgroup$
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
add a comment |
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1 Answer
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$begingroup$
I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X to Y$ is a closed continuous map.
- Then given $y in Y$ and an open $U subseteq X$ with $f^{-1} ( y ) subseteq U$ there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U$.
- Then $overline{ f(A) } = f ( overline{A} )$ for all $A subseteq X$.
(In fact, both of the above are equivalent characterizations of closed mappings.)
Pick $y in Y$ and an open neighbourhood $V subseteq Y$ of $y$. Then $K = f^{-1} ( y ) subseteq X$ is compact and $U = f^{-1} ( V ) subseteq X$ is open and $K subseteq U$.
By regularity of $X$ for each $x in K$ there is an open $U_x subseteq X$ with $x in U_x subseteq overline{U_x} subseteq U$. As ${ U_x : x in K }$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , ldots , x_n in K$ such that $K subseteq U_{x_1} cup cdots cup U_{x_n}$. Note that $overline{U_{x_1} cup cdots cup U_{x_n}} = overline{U_{x_1}} cup cdots cup overline{U_{x_n}} subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U_{x_1} cup cdots cup U_{x_n}$.
Using Fact(2) above (and the surjectivity of $f$) it follows that $$overline{W} = overline{f(f^{-1} (W)} = f ( overline{f^{-1}(W) } ) subseteq f ( overline{ U_{x_1} cup cdots cup U_{x_n} } ) subseteq f ( U ) = V.$$
That is, we have found an open $W subseteq Y$ such that $y in W subseteq overline{W} subseteq V$.
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
$endgroup$
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
add a comment |
$begingroup$
I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X to Y$ is a closed continuous map.
- Then given $y in Y$ and an open $U subseteq X$ with $f^{-1} ( y ) subseteq U$ there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U$.
- Then $overline{ f(A) } = f ( overline{A} )$ for all $A subseteq X$.
(In fact, both of the above are equivalent characterizations of closed mappings.)
Pick $y in Y$ and an open neighbourhood $V subseteq Y$ of $y$. Then $K = f^{-1} ( y ) subseteq X$ is compact and $U = f^{-1} ( V ) subseteq X$ is open and $K subseteq U$.
By regularity of $X$ for each $x in K$ there is an open $U_x subseteq X$ with $x in U_x subseteq overline{U_x} subseteq U$. As ${ U_x : x in K }$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , ldots , x_n in K$ such that $K subseteq U_{x_1} cup cdots cup U_{x_n}$. Note that $overline{U_{x_1} cup cdots cup U_{x_n}} = overline{U_{x_1}} cup cdots cup overline{U_{x_n}} subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U_{x_1} cup cdots cup U_{x_n}$.
Using Fact(2) above (and the surjectivity of $f$) it follows that $$overline{W} = overline{f(f^{-1} (W)} = f ( overline{f^{-1}(W) } ) subseteq f ( overline{ U_{x_1} cup cdots cup U_{x_n} } ) subseteq f ( U ) = V.$$
That is, we have found an open $W subseteq Y$ such that $y in W subseteq overline{W} subseteq V$.
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
$endgroup$
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
add a comment |
$begingroup$
I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X to Y$ is a closed continuous map.
- Then given $y in Y$ and an open $U subseteq X$ with $f^{-1} ( y ) subseteq U$ there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U$.
- Then $overline{ f(A) } = f ( overline{A} )$ for all $A subseteq X$.
(In fact, both of the above are equivalent characterizations of closed mappings.)
Pick $y in Y$ and an open neighbourhood $V subseteq Y$ of $y$. Then $K = f^{-1} ( y ) subseteq X$ is compact and $U = f^{-1} ( V ) subseteq X$ is open and $K subseteq U$.
By regularity of $X$ for each $x in K$ there is an open $U_x subseteq X$ with $x in U_x subseteq overline{U_x} subseteq U$. As ${ U_x : x in K }$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , ldots , x_n in K$ such that $K subseteq U_{x_1} cup cdots cup U_{x_n}$. Note that $overline{U_{x_1} cup cdots cup U_{x_n}} = overline{U_{x_1}} cup cdots cup overline{U_{x_n}} subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U_{x_1} cup cdots cup U_{x_n}$.
Using Fact(2) above (and the surjectivity of $f$) it follows that $$overline{W} = overline{f(f^{-1} (W)} = f ( overline{f^{-1}(W) } ) subseteq f ( overline{ U_{x_1} cup cdots cup U_{x_n} } ) subseteq f ( U ) = V.$$
That is, we have found an open $W subseteq Y$ such that $y in W subseteq overline{W} subseteq V$.
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
$endgroup$
I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X to Y$ is a closed continuous map.
- Then given $y in Y$ and an open $U subseteq X$ with $f^{-1} ( y ) subseteq U$ there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U$.
- Then $overline{ f(A) } = f ( overline{A} )$ for all $A subseteq X$.
(In fact, both of the above are equivalent characterizations of closed mappings.)
Pick $y in Y$ and an open neighbourhood $V subseteq Y$ of $y$. Then $K = f^{-1} ( y ) subseteq X$ is compact and $U = f^{-1} ( V ) subseteq X$ is open and $K subseteq U$.
By regularity of $X$ for each $x in K$ there is an open $U_x subseteq X$ with $x in U_x subseteq overline{U_x} subseteq U$. As ${ U_x : x in K }$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , ldots , x_n in K$ such that $K subseteq U_{x_1} cup cdots cup U_{x_n}$. Note that $overline{U_{x_1} cup cdots cup U_{x_n}} = overline{U_{x_1}} cup cdots cup overline{U_{x_n}} subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W subseteq Y$ with $y in W$ and $f^{-1} ( W ) subseteq U_{x_1} cup cdots cup U_{x_n}$.
Using Fact(2) above (and the surjectivity of $f$) it follows that $$overline{W} = overline{f(f^{-1} (W)} = f ( overline{f^{-1}(W) } ) subseteq f ( overline{ U_{x_1} cup cdots cup U_{x_n} } ) subseteq f ( U ) = V.$$
That is, we have found an open $W subseteq Y$ such that $y in W subseteq overline{W} subseteq V$.
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
answered Nov 29 '18 at 3:49
stochastic randomnessstochastic randomness
40017
40017
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
add a comment |
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
Is your first fact not generally true for continuous maps?
$endgroup$
– IntegrateThis
Nov 29 '18 at 5:31
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
$begingroup$
@IntegrateThis Consider $f(x) = sin ( pi x )$. Then $f^{-1}(0) = mathbb{Z}$. Let $U = bigcup_{n in mathbb{Z}} ( n - 2^{-|n|} , n + 2^{-|n|} )$. If $W$ is any open neighbourhood of $0$, then there is an $n > 1$ such that $( - 2^{-n} , 2^{-n} ) subseteq W$. Note that for $m > n$ we have that $m+2^{-m} in f^{-1}(W)$ but is not in $U$, so $f^{-1}(W) nsubseteq U$.
$endgroup$
– stochastic randomness
Nov 29 '18 at 5:49
add a comment |
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