Mixing
Continuous maps between compact Hausdorff spaces and their induced maps on their space of continuous...
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Let $X,Y$ be two compact Hausdorff spaces and let $ alpha: X to Y$ be continuous onto map. Let $alpha^{*}$ be the induced map from $C(Y)$ to $C(X)$ by mapping any $f in C(Y)$ to $f circ alpha in C(X)$.
First, is it true that the map $alpha^{*}$ is surjective?
Second, if my first proposition is not always true, then if we substitute "Compact Hausdorff space" with "extremally disconnected compact Hausdorff space", is my first proposition always true? Why?
Thanks.
general-topology functional-analysis compactness
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be two compact Hausdorff spaces and let $ alpha: X to Y$ be continuous onto map. Let $alpha^{*}$ be the induced map from $C(Y)$ to $C(X)$ by mapping any $f in C(Y)$ to $f circ alpha in C(X)$.
First, is it true that the map $alpha^{*}$ is surjective?
Second, if my first proposition is not always true, then if we substitute "Compact Hausdorff space" with "extremally disconnected compact Hausdorff space", is my first proposition always true? Why?
Thanks.
general-topology functional-analysis compactness
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
$endgroup$
2
$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
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And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38
add a comment |
$begingroup$
Let $X,Y$ be two compact Hausdorff spaces and let $ alpha: X to Y$ be continuous onto map. Let $alpha^{*}$ be the induced map from $C(Y)$ to $C(X)$ by mapping any $f in C(Y)$ to $f circ alpha in C(X)$.
First, is it true that the map $alpha^{*}$ is surjective?
Second, if my first proposition is not always true, then if we substitute "Compact Hausdorff space" with "extremally disconnected compact Hausdorff space", is my first proposition always true? Why?
Thanks.
general-topology functional-analysis compactness
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
$endgroup$
Let $X,Y$ be two compact Hausdorff spaces and let $ alpha: X to Y$ be continuous onto map. Let $alpha^{*}$ be the induced map from $C(Y)$ to $C(X)$ by mapping any $f in C(Y)$ to $f circ alpha in C(X)$.
First, is it true that the map $alpha^{*}$ is surjective?
Second, if my first proposition is not always true, then if we substitute "Compact Hausdorff space" with "extremally disconnected compact Hausdorff space", is my first proposition always true? Why?
Thanks.
general-topology functional-analysis compactness
general-topology functional-analysis compactness
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
asked Jan 26 at 15:51
AmirhosseinAmirhossein
817
817
2
$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
$begingroup$
And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38
add a comment |
2
$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
$begingroup$
And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38
2
2
$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
$begingroup$
And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38
add a comment |
1 Answer
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Suppose that $Y$ is a point and $X$ is $S^n$ for example, $fcirc alpha$ is always constant, there are non constant functions on $S^n$.
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
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$begingroup$
Suppose that $Y$ is a point and $X$ is $S^n$ for example, $fcirc alpha$ is always constant, there are non constant functions on $S^n$.
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
add a comment |
$begingroup$
Suppose that $Y$ is a point and $X$ is $S^n$ for example, $fcirc alpha$ is always constant, there are non constant functions on $S^n$.
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
add a comment |
$begingroup$
Suppose that $Y$ is a point and $X$ is $S^n$ for example, $fcirc alpha$ is always constant, there are non constant functions on $S^n$.
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
$endgroup$
Suppose that $Y$ is a point and $X$ is $S^n$ for example, $fcirc alpha$ is always constant, there are non constant functions on $S^n$.
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
answered Jan 26 at 15:57
Tsemo AristideTsemo Aristide
59.7k11446
59.7k11446
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$begingroup$
$alpha^*$ will never be surjective unless $alpha$ is injective. Just pick some point $y$ with more than one preimage, and construct a function which is 1 on one of those preimage points and 0 on another: this cannot be the pullback of a function on $Y$.
$endgroup$
– user98602
Jan 26 at 15:58
$begingroup$
And if $alpha$ is injective it is a homeomorphism from $X$ to $alpha[X]$ and we can show that $alpha^ast$ is surjective.
$endgroup$
– Henno Brandsma
Jan 26 at 16:21
$begingroup$
@MikeMiller I was reading an article in American mathematical monthly, in that was said if $alpha: X to Y$ is continuous onto map from an extremally disconnected compact Hausdorff space to a compact Hausdorff space then $alpha^{*}$ is a norm isomorphism between their space of functions. :-?
$endgroup$
– Amirhossein
Jan 27 at 6:38