Proving Tychonoff's theorem using the Stone-Cech compactification

Question

Let $X_i, \ i \in I$ be a collection of compact spaces. Consider the product space $X= \prod_i X_i$. Let $βX$ be the Stone-Cech compactification of $X$. For $i \in I$, let $π_i \colon X \to X_i$ be the i-th projection. Since each $X_i$ is compact, we can continuously extend $π_i$ onto $β X$. Denote that (unique) extension by $\overline{π}_i$. Now, consider the map $Φ \colon βX \to X$ defined by $Φ(p) = (\overline{π}_i p)_{i \in I}$. I want to show that $Φ$ is continuous, and as a result $X$ is compact. In order to do that, it suffices to show that each projection of $Φ$, $π_i \circ Φ \colon βX \to X_i$ is continuous. To this end, let $j \in I$. I noticed that $ π_j \circ Φ $ coincides with $\overline π_j$ on $X$ (here I view $X$ as a subset of $βX$). Indeed, if $x \in X$ then $$π_j \circ Φ(x) = π_j (\overline π_i x)_{i \in I}= π_j (π_i x) _{i \in I}= π_j(x) = \overline π_j (x) $$ since $\overline {π}_j = π_j$ on $X$. Hence, in order to show that $π_j \circ Φ$ is continuous it is both necessary and sufficient to prove that $π_j \circ Φ =\overline π_j$. Here is where I'm stuck. Any suggestions?

Answer 1

In a product space $X=\prod_{i \in I} X_i$, any map $f: Y \to X$ is continuous iff for all $i$, $\pi_i \circ f$ is continuous. This is well-known and classical (and follows from the definition of the product topology).

For your $\Phi$ it's true by definition that $\pi_i \circ \Phi = \overline{\pi_i}$ and so is indeed continuous. I don't see the problem.

The identity holds as for all $x \in X$ $\pi_i(\Phi(x)) = \overline{\pi}_i(x)$ and so we have equality of the functions $\pi_i \circ \Phi$ and $\overline{\pi}_i$, as defined on $X$.

You still have to show $\Phi$ is surjective to show compactness but for that it suffices to note that $\Phi\restriction_X = \textrm{id}_X$, e.g. (considering, as you do too that $X \subseteq \beta X$ which is somewhat inaccurate).