# If $f: X \to Y$, when do we have $\beta Y \supset \overline{f(X)} = \beta X$?

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?

-