The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.
We were going over separation axioms in class when assigned the following problem.
Given $X,Y,Z$ are $\mathscr{T}_{3\frac{1}{2}}$ and $X\xrightarrow{f}Y\xrightarrow{g}Z$; prove that $\beta{(g\circ f)}$=$\beta g\circ\beta f$.
What I understand:
- Suppose $X$ and $Y$ are completely regular and $f:X\xrightarrow{cts.}Y$. Then there is exactly one $\beta f:\beta X\xrightarrow{cts.}\beta Y$ such that $\beta f\circ e_x=e_y\circ f$. For each $g\in \mathcal{C}_y=\{h:Y\rightarrow[0,1]:h$ is cts. $\}$, we have $g\circ f\in\mathcal{C}_x=\{h:X\rightarrow[0,1]:h$ is cts $\}$, so let $\beta f(\{r_h\}_{h\in\mathcal{C}_x})=\{r_{g\circ f}\}_{g\in\mathcal{C}_y}$.
I have a rough idea:
- $\{r_{\phi}\}_{\phi\in\mathcal{C}_x}\in \beta X$
- $\{r_{\phi}\}_{\phi\in\mathcal{C}_x} \xrightarrow{\beta f}\{r_{\phi'\circ f}\}_{\phi'\in\mathcal{C}_y}\xrightarrow{\beta g} \{S_{\phi"\circ g}\}_{\phi"\in\mathcal{C}_z}$. Note I made: $S_{\phi'}=r_{\phi'\circ f}$
- $\{r_{\phi}\}_{\phi\in\mathcal{C}_x} \xrightarrow{\beta (g\circ f)}\{r_{\phi"\circ g \circ f}\}_{\phi"\in\mathcal{C}_z}$
I am having a rather difficult time proving $\{r_{\phi"\circ g \circ f}\}_{\phi"\in\mathcal{C}_z}=\{S_{\phi"\circ g}\}_{\phi"\in\mathcal{C}_z}$. Any suggestions? Any improvements?
I sincerely thank you for taking the time to read this question. If you need me to clarify, let me know. I greatly appreciate any assistance you may provide.
HINT: Don’t look at the details of the embeddings of $X,Y$, and $Z$ in their Čech-Stone compactifications. You know that $\beta g\circ\beta f$ is a continuous map from $\beta X$ to $\beta Z$, so all you have to show is that
$$\beta g\circ\beta f\circ e_x=e_z\circ g\circ f\;:$$
by uniqueness of $\beta(g\circ f)$ that tells you that $\beta g\circ\beta f=\beta(g\circ f)$. You know that $$\beta f\circ e_x=e_y\circ f\;,$$ so ...