If $f: X to Y$, when do we have $beta Y supset overline{f(X)} = beta X$?
$begingroup$
Suppose that $X$ and $Y$ are Tychonoff spaces, denote by $beta X$ and $beta Y$ their Stone-Čech compactifications and let $f:Xto Y$ be a continuous map.
Using the embedding $Yhookrightarrowbeta Y$ we can regard $f(X)$ as a subset of $beta Y$ and therefore take its closure $overline{f(X)} subset beta Y$.
Under which conditions do we have $overline{f(X)} = beta X$?
Let us regard $beta X$ as the Gelfand space to $C_b(X)$, i.e., for every element $x in beta X$ there is a net $(x_i) subset X$, such that we can regard $x$ as the multiplicative functional $varphi mapsto lim_i varphi(x_i)$ on $C_b(X)$. Analogously with $beta Y$.
Then it seems that there is a natural map $j: beta X to beta Y$ by pulling back bounded continuous functions from $Y$ to $X$ via the map $f$, i.e., $j(x)(varphi) := x(f^ast varphi)$, where $varphi in C_b(Y)$ and $x in beta X$. If $(x_i) subset X$ is a net with $x_i to x$ in $beta X$, then we have $x(f^ast varphi) = lim_i varphi(f(x_i))$. So $f(x_i) to j(x)$ in $beta Y$ and that means $j(beta X) subset overline{f(X)}$.
Now we want to define a map $i: beta Y supset overline{f(X)} to beta X$ with $i = j^{-1}$. The only way I see is to assume that $Y$ is a normal space, $f$ is an embedding and $f(X) subset Y$ is a closed subset. Then, given a function $psi in C_b(X)$, we can "push it forward" via $f$ to a continuous function $f_ast psi$ on $f(X) subset Y$ and extend it arbitrarily to the whole of $Y$ via the Tietze extension theorem. Call this extended function $widetilde{f_ast psi} in C_b(Y)$. Then we define $i(y)(psi) := y(widetilde{f_ast psi})$ for $y in overline{f(X)} subset beta Y$ and $psi in C_b(X)$. Since $y in overline{f(X)}$, this map $i$ is well-defined, i.e., independent of the extension of $f_ast psi$ to $widetilde{f_ast psi}$.
Is the above reasoning right, i.e., do we have $overline{f(X)} = beta X$ via the above maps $j$ and $i$ under the conditions that $Y$ is normal, $f$ an embedding and $f(X) subset Y$ closed?
What about the converse? Assume that we have $overline{f(X)} = beta X$. Can we conclude that $f$ must be an embedding or that $f(X) subset Y$ is closed or that $Y$ must be normal?
general-topology compactness compactification
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
$endgroup$
add a comment |
$begingroup$
Suppose that $X$ and $Y$ are Tychonoff spaces, denote by $beta X$ and $beta Y$ their Stone-Čech compactifications and let $f:Xto Y$ be a continuous map.
Using the embedding $Yhookrightarrowbeta Y$ we can regard $f(X)$ as a subset of $beta Y$ and therefore take its closure $overline{f(X)} subset beta Y$.
Under which conditions do we have $overline{f(X)} = beta X$?
Let us regard $beta X$ as the Gelfand space to $C_b(X)$, i.e., for every element $x in beta X$ there is a net $(x_i) subset X$, such that we can regard $x$ as the multiplicative functional $varphi mapsto lim_i varphi(x_i)$ on $C_b(X)$. Analogously with $beta Y$.
Then it seems that there is a natural map $j: beta X to beta Y$ by pulling back bounded continuous functions from $Y$ to $X$ via the map $f$, i.e., $j(x)(varphi) := x(f^ast varphi)$, where $varphi in C_b(Y)$ and $x in beta X$. If $(x_i) subset X$ is a net with $x_i to x$ in $beta X$, then we have $x(f^ast varphi) = lim_i varphi(f(x_i))$. So $f(x_i) to j(x)$ in $beta Y$ and that means $j(beta X) subset overline{f(X)}$.
Now we want to define a map $i: beta Y supset overline{f(X)} to beta X$ with $i = j^{-1}$. The only way I see is to assume that $Y$ is a normal space, $f$ is an embedding and $f(X) subset Y$ is a closed subset. Then, given a function $psi in C_b(X)$, we can "push it forward" via $f$ to a continuous function $f_ast psi$ on $f(X) subset Y$ and extend it arbitrarily to the whole of $Y$ via the Tietze extension theorem. Call this extended function $widetilde{f_ast psi} in C_b(Y)$. Then we define $i(y)(psi) := y(widetilde{f_ast psi})$ for $y in overline{f(X)} subset beta Y$ and $psi in C_b(X)$. Since $y in overline{f(X)}$, this map $i$ is well-defined, i.e., independent of the extension of $f_ast psi$ to $widetilde{f_ast psi}$.
Is the above reasoning right, i.e., do we have $overline{f(X)} = beta X$ via the above maps $j$ and $i$ under the conditions that $Y$ is normal, $f$ an embedding and $f(X) subset Y$ closed?
What about the converse? Assume that we have $overline{f(X)} = beta X$. Can we conclude that $f$ must be an embedding or that $f(X) subset Y$ is closed or that $Y$ must be normal?
general-topology compactness compactification
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
$endgroup$
add a comment |
$begingroup$
Suppose that $X$ and $Y$ are Tychonoff spaces, denote by $beta X$ and $beta Y$ their Stone-Čech compactifications and let $f:Xto Y$ be a continuous map.
Using the embedding $Yhookrightarrowbeta Y$ we can regard $f(X)$ as a subset of $beta Y$ and therefore take its closure $overline{f(X)} subset beta Y$.
Under which conditions do we have $overline{f(X)} = beta X$?
Let us regard $beta X$ as the Gelfand space to $C_b(X)$, i.e., for every element $x in beta X$ there is a net $(x_i) subset X$, such that we can regard $x$ as the multiplicative functional $varphi mapsto lim_i varphi(x_i)$ on $C_b(X)$. Analogously with $beta Y$.
Then it seems that there is a natural map $j: beta X to beta Y$ by pulling back bounded continuous functions from $Y$ to $X$ via the map $f$, i.e., $j(x)(varphi) := x(f^ast varphi)$, where $varphi in C_b(Y)$ and $x in beta X$. If $(x_i) subset X$ is a net with $x_i to x$ in $beta X$, then we have $x(f^ast varphi) = lim_i varphi(f(x_i))$. So $f(x_i) to j(x)$ in $beta Y$ and that means $j(beta X) subset overline{f(X)}$.
Now we want to define a map $i: beta Y supset overline{f(X)} to beta X$ with $i = j^{-1}$. The only way I see is to assume that $Y$ is a normal space, $f$ is an embedding and $f(X) subset Y$ is a closed subset. Then, given a function $psi in C_b(X)$, we can "push it forward" via $f$ to a continuous function $f_ast psi$ on $f(X) subset Y$ and extend it arbitrarily to the whole of $Y$ via the Tietze extension theorem. Call this extended function $widetilde{f_ast psi} in C_b(Y)$. Then we define $i(y)(psi) := y(widetilde{f_ast psi})$ for $y in overline{f(X)} subset beta Y$ and $psi in C_b(X)$. Since $y in overline{f(X)}$, this map $i$ is well-defined, i.e., independent of the extension of $f_ast psi$ to $widetilde{f_ast psi}$.
Is the above reasoning right, i.e., do we have $overline{f(X)} = beta X$ via the above maps $j$ and $i$ under the conditions that $Y$ is normal, $f$ an embedding and $f(X) subset Y$ closed?
What about the converse? Assume that we have $overline{f(X)} = beta X$. Can we conclude that $f$ must be an embedding or that $f(X) subset Y$ is closed or that $Y$ must be normal?
general-topology compactness compactification
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
$endgroup$
Suppose that $X$ and $Y$ are Tychonoff spaces, denote by $beta X$ and $beta Y$ their Stone-Čech compactifications and let $f:Xto Y$ be a continuous map.
Using the embedding $Yhookrightarrowbeta Y$ we can regard $f(X)$ as a subset of $beta Y$ and therefore take its closure $overline{f(X)} subset beta Y$.
Under which conditions do we have $overline{f(X)} = beta X$?
Let us regard $beta X$ as the Gelfand space to $C_b(X)$, i.e., for every element $x in beta X$ there is a net $(x_i) subset X$, such that we can regard $x$ as the multiplicative functional $varphi mapsto lim_i varphi(x_i)$ on $C_b(X)$. Analogously with $beta Y$.
Then it seems that there is a natural map $j: beta X to beta Y$ by pulling back bounded continuous functions from $Y$ to $X$ via the map $f$, i.e., $j(x)(varphi) := x(f^ast varphi)$, where $varphi in C_b(Y)$ and $x in beta X$. If $(x_i) subset X$ is a net with $x_i to x$ in $beta X$, then we have $x(f^ast varphi) = lim_i varphi(f(x_i))$. So $f(x_i) to j(x)$ in $beta Y$ and that means $j(beta X) subset overline{f(X)}$.
Now we want to define a map $i: beta Y supset overline{f(X)} to beta X$ with $i = j^{-1}$. The only way I see is to assume that $Y$ is a normal space, $f$ is an embedding and $f(X) subset Y$ is a closed subset. Then, given a function $psi in C_b(X)$, we can "push it forward" via $f$ to a continuous function $f_ast psi$ on $f(X) subset Y$ and extend it arbitrarily to the whole of $Y$ via the Tietze extension theorem. Call this extended function $widetilde{f_ast psi} in C_b(Y)$. Then we define $i(y)(psi) := y(widetilde{f_ast psi})$ for $y in overline{f(X)} subset beta Y$ and $psi in C_b(X)$. Since $y in overline{f(X)}$, this map $i$ is well-defined, i.e., independent of the extension of $f_ast psi$ to $widetilde{f_ast psi}$.
Is the above reasoning right, i.e., do we have $overline{f(X)} = beta X$ via the above maps $j$ and $i$ under the conditions that $Y$ is normal, $f$ an embedding and $f(X) subset Y$ closed?
What about the converse? Assume that we have $overline{f(X)} = beta X$. Can we conclude that $f$ must be an embedding or that $f(X) subset Y$ is closed or that $Y$ must be normal?
general-topology compactness compactification
general-topology compactness compactification
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
edited Dec 6 '18 at 7:19
Alex Ravsky
42.4k32383
42.4k32383
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
asked Apr 22 '13 at 21:02
AlexEAlexE
1,130924
1,130924
add a comment |
add a comment |
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