Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $X$ and $Y$ are Tychonoff spaces, denote by $\beta X$ and $\beta Y$ their Stone-Čech compactifications and let $f:X\to Y$ be a continuous map.

Using the embedding $Y\hookrightarrow\beta Y$ 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?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.